pLEASE help best and right answer gets brainliest

Step-by-step explanation:

ANSWER:(2x+2,y)

We have to convert it to m/s

[tex]\boxed{\sf 1km/h=\dfrac{5}{18}m/s}[/tex]

[tex]\\ \sf\longmapsto 40km/h[/tex]

[tex]\\ \sf\longmapsto 40\times \dfrac{5}{18}[/tex]

[tex]\\ \sf\longmapsto \dfrac{200}{18}[/tex]

[tex]\\ \sf\longmapsto 11.1m/s[/tex]

km/h > m/s

when converting km/h to m/s all you need to do is divide by 3.6

and vise versa when converting m/s to km/h multiply by 3.6

so therefore,

40km/h > m/s

= 40 / 3.6

= 11.11 m/s (4sf)

Answer:

132 degrees

Step-by-step explanation:

Looking at angle A and angle B, they are alternate interior angles. That means they are congruent to one another. Knowing that, we can set up an equation A=B

We can now fill A and B with their given equations

5x-18=3x+42

Now we solve

2x=60

x=30

Now that we know x is 30, we can replace it in the equation for A

5x-18

5(30)-18

150-18

132 degrees

Answer:

132

Step-by-step explanation:

ANGLE A = ANGLE B

(INTERIOR ALTERNATE ANGLES)

5x - 18 = 3x  + 42

2x = 60

x = 30

angle a = 150 - 18

= 132

Answer:

i believe the answer is 5.3

Let

ATQ

[tex]\\ \sf\longmapsto x+4x-6=94[/tex]

[tex]\\ \sf\longmapsto 5x-6=94[/tex]

[tex]\\ \sf\longmapsto 5x=94+6[/tex]

[tex]\\ \sf\longmapsto 5x=100[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{100}{5}[/tex]

[tex]\\ \sf\longmapsto x=20[/tex]

Answer:

$20

Step-by-step explanation:

T is Ted

K is Katie

T+K=94.

T=4K-6

I mostly just tried a bunch of numbers.

To check:

T+20=94

T=74

74=4(20)-6

74=80-6

74=74

I hope this helps!

pls ❤ and give brainliest pls

Answer: Option D. will be the answer.

Explanation: The bowling scores for six persons have been given as 27, 142, 145, 146, 154, 162.

The most appropriate measure of the center of these scores will be the median.

Here median will be mean of 146 and 146 because number of persons are 6 which is an even number.

So there are two center scores those are 145 and 146 and median =  

Option D. will be the answer.

9514 1404 393

Answer:

  -8,257,536·u^5·v^10

Step-by-step explanation:

The expansion of (a +b)^n is ...

  (c0)a^nb^0 +(c1)a^(n-1)b^1 +(c2)a^(n-2)b^2 +... +(ck)a^(n-k)b^k +... +(cn)a^0b^n

Then the k-th term is (ck)a^(n-k)b^k, where k is counted from 0 to n.

The value of ck can be found using Pascal's triangle, or by the formula ...

  ck = n!/(k!(n-k)!) . . . . where x! is the factorial of x, the product of all positive integers less than or equal to x.

This expansion has 11 terms, so the middle one is the one for k=5. That term will be ...

  5th term = (10!/(5!(10-5)!)(2u)^(10-5)(-4v^2)^5

  = (252)(32u^5)(-1024v^10) = -8,257,536·u^5·v^10

Answer:

y=3/2x+5

The slope is 3/2 and the y-intercept is 5

Step-by-step explanation:

Solving for y will give us the slope and y-intercept

Isolate y

2y/2=10+3x/2

y=5+3/2x

The slope is 3/2 and the y-intercept is 5

Graph it by graphing (0,5) and using the slope (up 3 over 2) to put other points

Answer:

x=11

Step-by-step explanation:

Answer:

x = 11

Step-by-step explanation:

6x - 10 = 4(x+3)

6x - 10 = 4*x + 4*3

6x - 10 = 4x + 12

6x - 4x = 12 + 10

2x = 22

x = 22/2

x = 11

check:

6*11 - 10 = 4(11+3)

66 - 10 = 4*14 = 56

Step-by-step explanation:

A manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process, at most 7 chips each lot would be defective, but if something does go wrong, there could be far more defective chips. If something goes wrong with a given lot, they discard the entire lot. It would be prohibitively expensive to test every chip in every lot, so they want to make the decision of whether or not to discard a given lot on the basis of the number of defective chips in a simple random sample. They decide they can afford to test 100 chips from each lot. You are hired as their statistician.

There is a tradeoff between the cost of eroneously discarding a good lot, and the cost of warranty claims if a bad lot is sold. The next few problems refer to this scenario.

Problem 8. (Continues previous problem.) A type I error occurs if (Q12)

Problem 9. (Continues previous problem.) A type II error occurs if (Q13)

Problem 10. (Continues previous problem.) Under the null hypothesis, the number of defective chips in a simple random sample of size 100 has a (Q14) distribution, with parameters (Q15)

Problem 11. (Continues previous problem.) To have a chance of at most 2% of discarding a lot given that the lot is good, the test should reject if the number of defectives in the sample of size 100 is greater than or equal to (Q16)

Problem 12. (Continues previous problem.) In that case, the chance of rejecting the lot if it really has 50 defective chips is (Q17)

Problem 13. (Continues previous problem.) In the long run, the fraction of lots with 7 defectives that will get discarded erroneously by this test is (Q18)

Problem 14. (Continues previous problem.) The smallest number of defectives in the lot for which this test has at least a 98% chance of correctly detecting that the lot was bad is (Q19)

(Continues previous problem.) Suppose that whether or not a lot is good is random, that the long-run fraction of lots that are good is 95%, and that whether each lot is good is independent of whether any other lot or lots are good. Assume that the sample drawn from a lot is independent of whether the lot is good or bad. To simplify the problem even more, assume that good lots contain exactly 7 defective chips, and that bad lots contain exactly 50 defective chips.

Problem 15. (Continues previous problem.) The number of lots the manufacturer has to produce to get one good lot that is not rejected by the test has a (Q20) distribution, with parameters (Q21)

Problem 16. (Continues previous problem.) The expected number of lots the manufacturer must make to get one good lot that is not rejected by the test is (Q22)

Problem 17. (Continues previous problem.) With this test and this mix of good and bad lots, among the lots that pass the test, the long-run fraction of lots that are actually bad is (Q23)

the models aren't given..

Answer: no models given

Step-by-step explanation:

what

Step-by-step explanation:

pretty sure its 25 percent

Answer:

25%

Step-by-step explanation:

if you take half of 50 it is 25 so all of it is used or 25%

Hope this helps <3 Comment if you want more thanks and be sure to give brainliest (4 left) <3

Answer:

The correct answer will be option b

Step-by-step explanation:

We know that x = rcos( θ ), and y = rsin( θ ), so let's rewrite this polar equation.

r = 4( x / r ) + 2( y / r ),

r = 4x / r + 2y / r,

r = 4x + 2y / r,

r / 1 = 4x + 2y / r ( Cross - Multiply )

4x + 2y = r²

We also know that r² = x² + y², so let's substitute.

x² + y² = 4x + 2y,

x² - 4x - 2y + y² = 0,

Circle Equation : ( x - 2 )² + ( y - 1 )² = ( √5 )²,

Solution : ( x - 2 )² + ( y - 1 )² = 5

Answer:

4.5cm

Step-by-step explanation:

Circumference = 2[tex]\pi[/tex]r

28.3=2[tex]\pi[/tex]r

28.3/2[tex]\pi[/tex]=r

4.456

The radius of the circle O is 4.5 cm.

A radius is a measure of distance from the center of any circular object to its outermost edge or boundary.

Given that, a circle having a circumference of approximately 28.3 cm, we need to find its radius,

So, Circumference = 2 π × radius

2 π × radius = 28.3

Radius = 28.3 / 2π

Radius = 28.3 / 6.28

Radius = 4.5

Hence, the radius of the circle O is 4.5 cm.

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Answer:

weight =48.71228786pounds

Step-by-step explanation:

[tex]w = \frac{k}{ {d}^{2} } \\ 50 = \frac{k}{ {3960}^{2} } \\ \\ k = 50 \times {3960}^{2} \\ k = 50 \times 15681600 \\ k = 784080000 \\ \\ w = \frac{784080000}{ {d}^{2} } \\ w = \frac{784080000}{16120225} \\ \\ w = 48.71228786 \\ w = 48.7pounds[/tex]

If a body weighs 50 pounds when it is 3,960 miles from Earth's center, it would weigh approximately 48.547 pounds if it were 4,015 miles from Earth's center, according to the inverse square law formula.

We know the inverse square law formula:

W₁ / W₂ = D²₂ / D²₁

Where W₁ is the weight of the body at the initial distance D₁, and W₂ is the weight at the final distance D₂.

So we have,

W₁ = 50

D₁  = 3,960

D₂  = 4015

We know that the body weighs 50 pounds when it is 3,960 miles from Earth's center,

So we can plug in those values as follows:

50 / W₂ = (4,015)²/ (3,960)²

To solve for W₂, we can cross-multiply and simplify as follows:

W₂ = 50 x (3,960)² / (4,015)²

W₂ = 50 x 15,681,600 / 16,120,225

W₂ = 48.547 pounds (rounded to three decimal places)

Therefore, if the body were 4,015 miles from Earth's center, it would weigh approximately 48.547 pounds.

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HA

See In Triangle DEF and Triangle XZY

[tex]\because\begin{cases}\sf \angle E=\angle Z=90° \\ \sf \ FD\sim XY=Hypotenuse\end{cases}[/tex]

Hence

[tex]\sf \Delta DEF\sim \Delta XZY(Angle-Angle)[/tex]

The theorems that verify that Δ DEF ~ Δ XZY is AA theorem of similarity.

Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion.

Given that, two triangles, Δ DEF and Δ XZY, we need to find a theorem that will verify that, Δ DEF and Δ XZY are similar,

So, we have, ∠ X = 40°,

Therefore, ∠ Y = 90°-40° = 50°

Now, we get,

∠ Y = ∠ F = 50°

∠ E = ∠ Z = 90°

We know that,

if two pairs of corresponding angles are congruent, then the triangles are similar.

Therefore, Δ DEF ~ Δ XZY by AA rule

Hence, the theorems that verify that Δ DEF ~ Δ XZY is AA theorem of similarity.

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Answer:

Step-by-step explanation:

Given that :

population Mean = 15000

standard deviation= 1200

sample size n = 100

sample mean = 16000

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o : \mu = 15000 }\\ \\ \mathtt{H_1 : \mu > 15000}[/tex]

Using the standard normal z statistics

[tex]z = \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}[/tex]

[tex]z = \dfrac{16000 -15000}{\dfrac{1200 }{\sqrt{100}}}[/tex]

[tex]z = \dfrac{1000}{\dfrac{1200 }{10}}[/tex]

[tex]z = \dfrac{1000\times 10}{1200}[/tex]

z = 8.333

degree of freedom = n - 1 = 100 - 1 = 99

level of significance ∝ = 0.05

P - value from the z score = 0.00003

Decision Rule: since the p value is lesser than the level of significance, we reject the null hypothesis

Conclusion: There is sufficient evidence  that the Dean claim for his graduate students earn more than average salary of $15,000

Dean's Claim of Average Salary = 16000, ie greater than 15000 : is correct

Null Hypothesis [ H0 ] : Average Salary = 15000

Alternate Hypothesis [ H1 ] : Average Salary > 15000

Hypothesis is tested using t statistic.

t = ( x - u ) / ( s / √ n ) ; where -

x = sample mean , u = population mean , s = standard deviation, n = sample size

t = ( 16000 - 15000 ) / ( 1200 / √100 )

= 1000 / 120

t  {Calculated} = 8.33,

Degrees of Freedom = n - 1 = 100 = 1 = 99

Tabulated t 0.05 (one tail) , at degrees of freedom 99 = 1.664

As Calculated t value 8.33 > Tabulated t value 1.664 , So we reject the Null Hypothesis in favour of Alternate Hypothesis.

So, conclusion : Average Salary > 15000

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Answer:

10.9

Step-by-step explanation:

[tex]a^2+b^2=c^2[/tex]

[tex]9.1^2+6^2=c^2[/tex]

[tex]82.81+36=c^2[/tex]

[tex]c^2=118.81[/tex]

[tex]c=\sqrt{118.81} =10.9[/tex]

Part a)

There are 52 letters (26 lowercase and 26 uppercase), 10 digits, and 6 symbols. There are 52+10+6 = 68 different characters to choose from.

Adding up those subtotals gives

68^8+68^9+68^10+68^11+68^12 = 9.9207 * 10^21

different passwords possible.

======================================================

Part b)

Let's find the number of passwords where we don't have a special symbol

There are 52+10 = 62 different characters to pick from

Adding those subtotals gives

62^8+62^9+62^10+62^11+62^12 = 3.2792 * 10^21

different passwords where we do not have a special character. Subtract this from the answer in part a) above

( 9.9207 * 10^21)  - (3.2792 * 10^21) = 6.6415 * 10^21

which represents the number of passwords where we have one or more character that is a special symbol. I'm using the idea that we either have a password with no symbols, or we have a password with at least one symbol. Adding up those two cases leads to the total number of passwords possible.

======================================================

Part c)

The answer from part a) was roughly 9.9207 * 10^21

It will take about 9.9207 * 10^21  nanoseconds to try every possible password from part a).

Divide 9.9207 * 10^21  over 1*10^9 to convert to seconds

(9.9207 * 10^21 )/(1*10^9) = 9,920,700,000,000

This number is 9.9 trillion roughly.

It will take about 9.9 trillion seconds to try every password, if you try a password per second.

------

To convert to hours, divide by 3600 and you should get

(9,920,700,000,000)/3600 = 2,755,750,000

So it will take about 2,755,750,000 hours to try all the passwords.

------

Divide by 24 to convert to days

(2,755,750,000)/24= 114,822,916.666667

which rounds to 114,822,917

So it will take roughly 114,822,917 days to try all the passwords.

------

Then divide that over 365 to convert to years

314,583.334246576

which rounds to 314,583

It will take roughly 314,583 years to try all the passwords

------------------------------

All values are approximate, and are roughly equivalent to one another.

A) 9,920,671,339,261,325,541,376 different passwords are available for this computer system.

B) 875,353,353,464,234,606,592 of these passwords contain at least one occurrence of at least one of the six special characters.

C) It would take 314,582.42 years for a hacker to try every possible password.

To determine how many different passwords are available for this computer system; how many of these passwords contain at least one occurrence of at least one of the six special characters; and how long it takes a hacker to try every possible password, assuming that it takes one nanosecond for a hacker to check each possible password, the following calculations must be performed:

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The triangle (call it T ) has base and height 4, so its area is 1/2*4*4 = 8. Then the joint density function is

[tex]f_{X,Y}(x,y)=\begin{cases}\frac18&\text{for }(x,y)\in T\\0&\text{otherwise}\end{cases}[/tex]

where T is the set

[tex]T=\{(x,y)\mid 0\le x\le4\land0\le y\le4-x\}[/tex]

(a) I've attached an image of the integration region.

[tex]P(X<3,Y<3)=\displaystyle\int_0^1\int_0^3f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx+\int_1^3\int_0^{4-x}f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\frac12[/tex]

(b) X and Y are independent if the joint distribution is equal to the product of their marginal distributions.

Get the marginal distributions of one random variable by integrating the joint density over all values of the other variable:

[tex]f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^{4-x}\frac{\mathrm dy}8=\begin{cases}\frac{4-x}8&\text{for }0\le x\le4\\0&\text{otherwise}\end{cases}[/tex]

[tex]f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^{4-y}\frac{\mathrm dx}8=\begin{cases}\frac{4-y}8&\text{for }0\le y\le4\\0&\text{otherwise}\end{cases}[/tex]

Clearly, [tex]f_{X,Y}(x,y)\neq f_X(x)f_Y(y)[/tex], so they are not independent.

Answer:

2.34

Step-by-step explanation:

A pharmacy purchased 550 products over a period of 3 months

The average inventory was 235 products during the period of 3 months

Therefore, the inventory turnover rate for this period can be calculated as follows

= 550/235

= 2.34

Hence the inventory turnover rate for this period is 2.34

Answer:

4

Step-by-step explanation:

Plug in 3 as x in the expression:

10 - 2x

10 - 2(3)

10 - 6

= 4

Answer:

4

Step-by-step explanation:

10 - 2x

Let x =3

10 -2(3)

10 -6

4

Answer:

C but see below.

Step-by-step explanation:

If I'm reading this correctly, you mean 31/24. It really can't be much else.  The sine and cosine are both incorrect because both involve the hypotenuse which must be calculated in order for them to be considered. In addition 31/24 is greater than one which is impossible for both the Sine and the Cosine.

That leaves K and D

Tan(D) = 24/31 which is not an option.

That leaves C.

tan(K) = 31/24 which is what you have to choose. If your choice is not written this way, then there is no answer.

Answer:

The answer to this problem is C. tanK=3124

Answer:

6

Step-by-step explanation:

Let a = -1, b = 2.5, and c = 3.

-1 + 2(3² - (2.5 + 3))

-1 + 2(9 - 5.5)

-1 + 2(3.5)

-1 + 7

6

Answer:

AA

Step-by-step explanation:

See In Triangle ABC and Triangle STU

[tex]\because\begin{cases}\sf \angle A=\angle S=90° \\ \sf \angle B=\angle T=31°\end{cases}[/tex]

Hence

[tex]\sf \Delta ABC~\Delta STU(Angle-Angle)[/tex]

By AA similarity  triangle ABC is similar to triangle SUT. Therefore, option A is the correct answer.

Two triangles are similar if the angles are the same size or the corresponding sides are in the same ratio. Either of these conditions will prove two triangles are similar.

In the given triangle ABC, ∠C=180°-90°-31°

∠C=59°

In the given triangle SUT, ∠U=180°-90°-31°

∠U=59°

Here, ∠B=∠T (Given)

∠C=∠U (Obtained using angle sum property of a triangle)

So, by AA similarity ΔABC is similar to ΔSUT.

Therefore, option A is the correct answer.

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Answer:

The answer would be D. DE

Step-by-step explanation:

same prob

Congruent triangles are exact same triangles, but they might be placed at different positions. The correct option is D.

Suppose it is given that two triangles ΔABC ≅ ΔDEF

Then that means ΔABC and ΔDEF are congruent. Congruent triangles are exact same triangles, but they might be placed at different positions.

The order in which the congruency is written matters.

For ΔABC ≅ ΔDEF, we have all of their corresponding elements like angle and sides congruent.

Thus, we get:

[tex]\rm m\angle A = m\angle D \: or \: \: \angle A \cong \angle D \angle B = \angle E\\\\\rm m\angle B = m\angle E \: or \: \: \angle B \cong \angle E \\\\\rm m\angle C = m\angle F \: or \: \: \angle C \cong \angle F \\\\\rm |AB| = |DE| \: \: or \: \: AB \cong DE\\\\\rm |AC| = |DF| \: \: or \: \: AC \cong DF\\\\\rm |BC| = |EF| \: \: or \: \: BC \cong EF[/tex]

(|AB| denotes the length of line segment AB, and so on for others).

Given that ΔABC ≅ ΔDEF. Therefore, the given sentence can be completed as AB ≅ ΔDE.

Hence, the side AB ≅ ΔDE

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Answer:

a) No sufficient evidence to support the claim that the two machines produce rods with different mean diameters.

b) P-value is 0.80

c)  −0.3939 <μ< 0.4939

Step-by-step explanation:

Given Data:

sample sizes

n1 = 15

n2 = 17

sample means:

x1 = 8.73

x2 = 8.68

sample variances:

s1² = 0.35

s2² = 0.40

Hypothesis:

H₀ : μ₁ = μ₂

H₁ :  μ₁ ≠ μ₂

Compute the pooled standard deviation:

[tex]s_{p} = \sqrt{\frac{(n_{1} - 1)s_{1}^{2} + (n_{2} - 1)s_{2}^{2}}{n_{1} +n_{2} -2} }[/tex]

    [tex]= \sqrt{\frac{(15-1)0.35+(17-1)0.40}{15+7-2}}[/tex]

    [tex]= \sqrt{\frac{(14)0.35+(16)0.40}{30}}[/tex]

 [tex]= \sqrt{\frac{4.9+6.4}{30}}[/tex]

 [tex]= \sqrt{\frac{11.3}{30}}[/tex]

[tex]= \sqrt{0.376667}[/tex]

= 0.613732

= 0.6137

Compute the test statistic:

[tex]t = \frac{x_{1} -x_{2} }{s_{p} \sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]

[tex]= \frac{8.73-8.68}{0.6137\sqrt{\frac{1}{15}+\frac{1}{17} } }[/tex]

[tex]= \frac{0.05}{0.6137\sqrt{0.06667+0.05882} } }[/tex]

[tex]= \frac{0.05}{0.6137\sqrt{0.12549} } }[/tex]

[tex]= \frac{0.05}{0.6137(0.354246)} } }[/tex]

[tex]= \frac{0.05}{0.6137(0.354246)} } }[/tex]

= 0.05 / 0.217401

= 0.22999

t = 0.230

Compute degree of freedom:

df = n1 + n2 -2 = 15 + 17 - 2 = 30

Compute the P-value from table using df = 30

P > 2 * 0.40 = 0.80

P > 0.05 ⇒ Fail to reject H₀

Null hypothesis is rejected when P-value is less than or equals to level of significance. But here the P-value = 0.80 and level of significance = 0.05. So P-value is greater than significance level. Hence there is not sufficient evidence to support the claim that population means are different.

Construct a 95% confidence interval for the difference in mean rod diameter:

confidence = c = 95% = 0.95

α = 1 - c

  = 1 - 0.95

α = 0.05

Compute degree of freedom:

df = n1 + n2 -2 = 15 + 17 - 2 = 30

Compute [tex]t_{\alpha /2}[/tex] with df = 30 using table:

t₀.₀₂₅ = 2.042

Compute confidence interval:

= [tex](x_{1}-x_{2})-t_{\alpha/2} ( s_{p} )\sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } }[/tex]

= (8.73 - 8.68) -  2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]

= 0.05 - 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]

= 0.05 - 1.253175 [tex]\sqrt{0.06667+0.05882} } }[/tex]

= 0.05 - 1.253175 [tex]\sqrt{0.12549} } }[/tex]

= 0.05 - 1.253175 (0.35424))

= 0.05 - 0.443925

= −0.393925

= −0.3939

[tex](x_{1}-x_{2})+t_{\alpha/2} ( s_{p} )\sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } }[/tex]

= (8.73 - 8.68) +  2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]

= 0.05 + 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]

= 0.05 + 1.253175 [tex]\sqrt{0.06667+0.05882} } }[/tex]

= 0.05 + 1.253175 [tex]\sqrt{0.12549} } }[/tex]

= 0.05 + 1.253175 (0.35424))

= 0.05 + 0.443925

= 0.493925

= 0.4939

−0.3939 <μ₁ - μ₂< 0.4939

Answer:

The order, least to greatest, is:

Lemon, Cherry, Grape.

Step-by-step explanation:

Adding all these values up, we get to 1. This means that the smallest values will be the least likely and the highest values will be the most likely.

With the numbers 0.2, 0.16, and 0.64, we can sort these by value.

0.16 is the smallest.

0.2 is the next biggest

and 0.64 is the largest number.

So, the order is Lemon, Cherry, Grape.

Hope this helped!