Define limit and it's types.​

Answer:

x = 10

Step-by-step explanation:

[tex]x = 3 \times 2 + 4[/tex]

[tex]x = 6 + 4[/tex]

[tex]x = 10[/tex]

Answer:

1. add 1/2x to both sides

a. you want to combine the like terms. in this case, it is the x variable.

you are left with 7/6x = 5

2. multiply by 6/7

a. the reciprocal of 7/6 will cancel out the values

Answer:

1. Jan checks the weather.  It is 27 degrees outside.  Jan did chores for two hours.  After Jan was done, she checked the weather again.  The temperature had decreased 11 degrees.

2. (See screen shot below.)

Step-by-step explanation:

1.  It doesn't have to be as complicated as I made it.  You can just say that the weather started out with 27 degrees, and decreased later on.  Remember, decreased means subtracted and 27+(-11) is the same as 27-11 because when a + and - are together - always wins.  So no.1 wants you to say something got subtracted.

2. On the number line, make a dot at 29 because it said it was 29 degrees.  Then drag the dot at the number 29 to 13 because it said it decreased 16, so it is 19 minus 16 which is 13.

Answer:

$20

Step-by-step explanation:

Company A:

Buy 300 shares at $1.45 per share.

Sell 300 shares at $1.65 per share.

Profit: ($1.65 - $1.45) * 300 = $60

Company B:

Buy 200 shares at $1.20 per share.

Sell 200 shares at $1.10 per share.

Loss: ($1.20 - $1.10) * 200 = $20

Net profit:

$40 - $20 = $20

Answer:

Step-by-step explanation:

Profit on Company A =$(1.65−1.45)×300=$60.

Loss on Company B =$(1.20−1.10)×200=$20.

Therefore the total profit Jamal achieved was $60−$20=$40.

Answer:

Bro 1st expression is in simplest form and 2nd in simplest is 3/2

Step-by-step explanation:

If you like my answer than please mark me brainliest

Answer:

(5/2)*(9/6)

three can go into the fraction(9/6) giving(3/2)

(5/2)*(3/2)=15/4

mark as brainliest

Answer: [tex]\dfrac{8}{\pi}[/tex] .

Step-by-step explanation:

We know that the rate of change of function f(x) from x=a to x= b is given by :-

[tex]k=\dfrac{f(b)-f(a)}{b-a}[/tex]

The given points on graph  :  (0, -4) and (pi over 2, 0) and (pi, 4) and (3 pi over 2, 0) and (2 pi, -4).

The rate of change from x = 0 to x = pi over 2 will be :-

[tex]\dfrac{0-(-4)}{\dfrac{\pi}{2}-0}=\dfrac{4}{\dfrac{\pi}{2}}[/tex]     [By using points (0, -4) and (pi over 2, 0) ]

[tex]=\dfrac{8}{\pi}[/tex]

Hence, the rate of change from x = 0 to x = pi over 2 is [tex]\dfrac{8}{\pi}[/tex] .

Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477]  .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                             P.Q.  =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean = $390

            s = sample standard deviation = $120

            n = sample of students = 19

            [tex]\mu[/tex] = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-2.101 < [tex]t_1_8[/tex] < 2.101) = 0.95  {As the critical value of t at 18 degrees of

                                               freedom are -2.101 & 2.101 with P = 2.5%}  

P(-2.101 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.101) = 0.95

P( [tex]-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.101 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.101 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                        = [ [tex]\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }[/tex] , [tex]\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }[/tex] ]

                        = [$332.16, $447.84]

(A)  Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is [tex]Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }[/tex] would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is [tex]Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }[/tex]  would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                             P.Q.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion students who purchase their used textbooks = [tex]\frac{210}{500}[/tex] = 0.42    

            n = sample of students = 500

            p = population proportion

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5%

                                               level of significance are -2.58 & 2.58}  

P(-2.58 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 2.58) = 0.99

P( [tex]-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99

P( [tex]\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99

99% confidence interval for p = [ [tex]\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]

= [ [tex]0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }[/tex] , [tex]0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }[/tex] ]

= [0.363, 0.477]

Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477]  .

Answer:

-2,  (-5 is extraneous)

Step-by-step explanation:

assuming √(x+6) -4 =x

√(x+6) = x+4

square both side

x+6 = (x+4)^2

x+6 = x^2+8x+16

x^2+7x+10=0

(x+2)(x+5) = 0

x = -2, -5

put -5 back into org equation and it is not true

√(-5+6) -4 ≠ -5

Answer

Is it C I may have done my math wrong lol

Step-by-step explanation:

Answer:

[tex]t = 0.811\,s[/tex] contains the cheapest reference to sugar; [tex]t = 4.511\,s[/tex] contains the most expensive reference to sugar.

Step-by-step explanation:

Let be [tex]s(t) = -0.00003237\cdot t^{5} + 0.0009037\cdot t^{4}-0.008956\cdot t^{3}+0.03629\cdot t^{2}-0.04547\cdot t + 0.4778[/tex], the times when sugar is the cheapest and the most expensive (absolute minimum and maximum) are determined with the help of first and second derivatives of this function (First and Second Derivative Tests):

First Derivative Test

[tex]s'(t) = -0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547[/tex]

Let equalize the polynomial to zero and solve the resulting expression:

[tex]-0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547 = 0[/tex]

[tex]t_{1} \approx 9.511\,s[/tex], [tex]t_{2}\approx 7.431\,s[/tex], [tex]t_{3}\approx 4.511\,s[/tex] and [tex]t_{4}\approx 0.881\,s[/tex]

Second Derivative Test

[tex]s''(t) = -0.0006474\cdot t^{3}+0.0108444\cdot t^{2}-0.053736\cdot t+0.07258[/tex]

This function is now evaluated at each root found in the First Derivative section:

[tex]s''(9.511\,s) = -0.0006474\cdot (9.511\,s)^{3}+0.0108444\cdot (9.511\,s)^{2}-0.053736\cdot (9.511\,s)+0.07258[/tex]

[tex]s''(9.511\,s) = -0.015[/tex] (A maximum)

[tex]s''(7.431\,s) = -0.0006474\cdot (7.431\,s)^{3}+0.0108444\cdot (7.431\,s)^{2}-0.053736\cdot (7.431\,s)+0.07258[/tex]

[tex]s''(7.431\,s) = 6.440\times 10^{-3}[/tex] (A minimum)

[tex]s''(4.511\,s) = -0.0006474\cdot (4.511\,s)^{3}+0.0108444\cdot (4.511\,s)^{2}-0.053736\cdot (4.511\,s)+0.07258[/tex]

[tex]s''(4.511\,s) = -8.577\times 10^{-3}[/tex] (A maximum)

[tex]s''(0.811\,s) = -0.0006474\cdot (0.811\,s)^{3}+0.0108444\cdot (0.811\,s)^{2}-0.053736\cdot (0.811\,s)+0.07258[/tex]

[tex]s''(0.811\,s) = 0.036[/tex] (A minimum)

Each value is evaluated in order to determine when sugar was the cheapest and the most expensive:

Cheapest (Absolute minimum)

[tex]s(0.811\,s) = -0.00003237\cdot (0.811\,s)^{5}+0.0009037\cdot (0.811\,s)^{4}-0.008956\cdot (0.811\,s)^{3}+0.03629\cdot (0.811\,s)^{2}-0.04547\cdot (0.811\,s)+0.4778[/tex]

[tex]s(0.811\,s) = 0.460[/tex]

[tex]s(7.431\,s) = -0.00003237\cdot (7.431\,s)^{5}+0.0009037\cdot (7.431\,s)^{4}-0.008956\cdot (7.431\,s)^{3}+0.03629\cdot (7.431\,s)^{2}-0.04547\cdot (7.431\,s)+0.4778[/tex]

[tex]s(7.431\,s) = 0.491[/tex]

[tex]t = 0.811\,s[/tex] contains the cheapest reference to sugar.

Most expensive (Absolute maximum)

[tex]s(4.511\,s) = -0.00003237\cdot (4.511\,s)^{5}+0.0009037\cdot (4.511\,s)^{4}-0.008956\cdot (4.511\,s)^{3}+0.03629\cdot (4.511\,s)^{2}-0.04547\cdot (4.511\,s)+0.4778[/tex]

[tex]s(4.511\,s) = 0.503[/tex]

[tex]s(9.511\,s) = -0.00003237\cdot (9.511\,s)^{5}+0.0009037\cdot (9.511\,s)^{4}-0.008956\cdot (9.511\,s)^{3}+0.03629\cdot (9.511\,s)^{2}-0.04547\cdot (9.511\,s)+0.4778[/tex]

[tex]s(9.511\,s) = 0.498[/tex]

[tex]t = 4.511\,s[/tex] contains the most expensive reference to sugar.

The required values are,

[tex]t=0.881199[/tex] at the cheapest.

[tex]t=4.51081[/tex] at the most expensive.

A high point is called a maximum (plural maxima ). A low point is called a minimum (plural minima ).

Given equation is,

[tex]S(t) = -0.00003237t^5 + 0.0009037t^4- 0.008956t^3 + 0.03629t^2-0.04547t + 0.4778[/tex]

Differentiating the given equation we get,

[tex]S'(t)=-0.00003237\times 5t^4+0.0009037\times 4t^3-0.008956\times 3t^2+0.03629\times 2t-0.04547+0\\S'(t)=0\\-0.00003237\times 5t^4+0.0009037\times 4t^3-0.008956\times 3t^2+0.03629\times 2t-0.04547+0=0\\t=0.881199\\t=4.51081\\t=7.43087\\t=9.51137\\[/tex]

Now we can directly plug those fours values of t into given function S(t) to find which one gives max or minimum or you can also use the 2nd derivative test. Although that is not compulsory

[tex]t=0.881199,S(t)=0.46031095\\t=4.51081, S(t)=0.50278423\\t=7.43087, S(t)=0.49096762\\t=9.51137, S(t)=0.49832202\\[/tex]

We see that sugar is cheapest at [tex]t=0.881199[/tex] which is approx 1 and corresponds to the year [tex]1993+1=1994[/tex]

Similarly sugar is most expensive at [tex]t=4.51081[/tex] which is approx 5 and corresponds to year [tex]1993+5=1998[/tex]

Learn more about the topic Minimum or Maximum:

https://brainly.com/question/10359210

Answer:

Step-by-step explanation:

Distance walked by Arthur = 5/8 miles

Distance ride  by Jonathan = 8 times that of Arthur

it means that Distance rode on bus by Jonathan is 8 multiplied by Distance walked by Arthur

Distance rode on bus by Jonathan = 8 * Distance walked by Arthur

Distance rode on bus by Jonathan = 8 * 5/8 = 5 Miles Answer

y = a(x  + 1.5)^2 - 12.5

y intercept is (0,-8) so:-

-8 = a(0+1.5)^2 - 12.5

-8 = 2.25a - 12.5

a =  4.5/ 2.25 = 2

so we have  

y  = 2 ( x +1.5)^2 - 12.5

solving for x  when y = 0:-

(x + 1.5)^2 = 12.5/2 = 6.25

taking sqrt's  x + 1.5  = +/- 2.5

x =  -4,   1

so the x intercepts are (-4,0) and (1,0)

Answer:

y = –1∕4(x – 8)^2 – 1

Step-by-step explanation:

I took the exam and  got  it  right.

Answer:

B

Step-by-step explanation:

Since the power is negative, you automatically know it has to be a or b, because the only way it would be negative is if it was brought from the denominator to the numerator.

The answer is B, because the numerator of the power, is what is inside the square root, while the denominator is what is outside the square root.

Answer:

By finding LCM of 9 and 12 the answer is 36

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

Answer:

236°

Step-by-step explanation:

The circumference of a circle is 360° since <DOC is given as 44° and <COB is given as 80° and the center angles are equal to the arc it sees the the measure of arc DEB would be 360 - 44 - 80 = 236°

Is this question complete?

Answer:

D

Step-by-step explanation:

(b-2)(b-8)+4=0

b^2-10b+16+4=0

b^2-10b+20=0

Product of roots is given by c/a=20. So the answer is 20

[tex]\sqrt{16}\times\sqrt{12}[/tex]

$=\sqrt{4^2}\times\sqrt{2^2\cdot3}$

$=4\times2\sqrt3=8\sqrt3$

Answer:

The greater than symbols looks like this    >    , and the less than symbol looks like? <

Answer:

Greater than symbol: >

Less than symbol: <

Greater than or equal to symbol: ≥

Less than or equal to symbol: ≤

Equal symbol: =

In this case, you are answering with the greater than symbol as well as the less than symbol.

The greater than symbols looks like this > , and the less than symbol looks like < .

Answer:

The number of ways is  [tex]\left n} \atop {}} \right. P_r = 336[/tex]

Step-by-step explanation:

From the question we are told that

   The number of tickets are   [tex]n = 8[/tex]

    The number of finalist are [tex]r =3[/tex]

Generally the number of way by which this winners can be drawn and arrange in the order of   [tex]1^{st} , \ 2nd , \ 3rd[/tex]    is mathematically represented as

             [tex]\left n} \atop {}} \right. P_r = \frac{n\ !}{(n-r) !}[/tex]

substituting values

             [tex]\left n} \atop {}} \right. P_r = \frac{ 8!}{(8-3) !}[/tex]

           [tex]\left n} \atop {}} \right. P_r = \frac{ 8* 7*6*5*4*3*2*1}{ 5*4*3*2*1}[/tex]

           [tex]\left n} \atop {}} \right. P_r = 336[/tex]

If the scale factor is 30, then all you have to do is multiply each measurement by the scale factor. In this case, 4 · 30 = 120.

Answer:

[tex]\Large \boxed{\sf \bf \ \ y=\dfrac{1}{16}(a-3)^2-2 \ \ }[/tex]

Step-by-step explanation:

Hello, please consider the following.

When the parabola equation is like

[tex]y=a(x-h)^2+k[/tex]

The vertex is the point (h,k) and the focus is the point (h, k+1/(4a))

As the vertex is (3,-2) we can say that h = 3 and k = -2.

We need to find a.

The focus  is (3,2) so we can say.

[tex]2=-2+\dfrac{1}{4a}\\\\\text{*** We add 2. ***}\\\\\dfrac{1}{4a}=2+2=4\\\\\text{*** We multiply by 4a. ***}\\\\16a=1\\\\\text{*** We divide by 16. ***}\\\\a=\dfrac{1}{16}[/tex]

So an equation for the parabola is.

[tex]\large \boxed{\sf y=\dfrac{1}{16}(a-3)^2-2 }[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Step-by-step explanation:

I used an app called DESMOS It Is usually super helpful!!!

Answer:

8

Step-by-step explanation:

4×16=64

[tex] \sqrt{64 } = 8[/tex]

9514 1404 393

Answer:

  x = 5.0

Step-by-step explanation:

The tangent relation is helpful:

  Tan = Opposite/Adjacent

  tan(50°) = x/4.2

  x = 4.2·tan(50°) ≈ 5.0054 . . . . multiply by 4.2

  x ≈ 5.0

Answer:

1. It would take the car to get from Tucson to Phoenix 12 hours.

2. for the car to go around the equator it would take 637 hours if it is still travelling at 10km/hr.

hope this helps

Step-by-step explanation:

1. 120 km divided by 10 = 12 hours

Answer:

The value for the Chi -square  test statistics = 1.149

Step-by-step explanation:

The observed value Table can be shown better as:

Observed Value

Age            Favorable          Unfavorable            Neutral              Total

18-30             67                            24                       20                   111

30-45            50                            14                        16                    80

Over 45         69                           21                        26                   116

Total              186                         59                        62                   307

NOTE: when computing the question, in the third row and the second column, there is a mistake , the value is supposed to be 21 and not  41 because :

69 +21+ 26 will eventually give = 116

69 + 41 + 26 = 136  

With that error being fix , let's get started.

Expected Value

The expected value can be determined by using the formula:

[tex]Expected \ Value = \dfrac{ row \ total \times column \ total }{grand \ total }[/tex]

For 67; (111 * 186)/307 = 67.251

For 24 : (111 * 59)/307 = 21.332

For 20 : (111 * 62)/307 = 22.417

For 50 :(80*186)/307 = 48.469

For 14 : (80* 59)/307 = 15.375

For 16 : ( 0 * 62)/307 = 16.156

For 69 : (116 * 186)/307 = 70.280

For 21 : (116* 59)/307 = 22.293

For 26 : (116*62)/307 = 23.427

Expected Value :

Age            Favorable          Unfavorable            Neutral              Total

18-30         67.251                 21.332                     22.417                  111

30-45        48.469                15.375                     16.156                  80

Over 45    70.280                22.293                     23.427               116

Total              186                         59                        62                   307

The Chi - square test statistics = [tex]\dfrac{(observed \ value - Expected \ value)^2}{Expected \ value}[/tex]

For 67.251 : ( 67 - 67.251)²/67.251 = 0.0009

For 21.332 : ( 24 - 21.332)²/21.332 = 0.3337

For 22.417 : ( 20 - 22.417)²/ 22.417 = 0.2606

For 48.469 : ( 50 - 48.469)²/ 48.469 = 0.0484

For 15.375 : ( 14 - 15.375)²/ 15.375 = 0.1230

For 16.156 : ( 16 - 16.156)²/ 16.156 = 0.0015

For 70.280 : ( 69 - 70.280)²/ 70.280 = 0.0233

For 22.293 : ( 21 - 22.293)²/ 22.293 = 0.0750

For 23.427 : ( 26 - 23.427)²/ 23.427 = 0.2826

The chi square table is as follows:

Age            Favorable          Unfavorable            Neutral          Total

18-30          0.0009              0.3337                   0.2606           0.5952

30-45         0.0484               0.1230                   0.0015            0.1729

Over 45     0.0233               0.0750                  0.2826            0.3809

Total          0.0726                0.5317                   0.5447              1.149

The value for the Chi -square  test statistics = 1.149

Answer: a) 0.1095     b) 0.0095

Step-by-step explanation:

Given : The deck for a card game contains 30 cards.

10 are red, 10 yellow, 5 blue, and 1 green, and 4 are wild cards.

Each player is randomly dealt a five-card hand.

Number of ways to choose 5 cards out of 30 = [tex]C(30,5)=\dfrac{30!}{5!25!}=142506[/tex]

a)  Cards other than wild card = 30-4=26

Number of ways to choose exactly two wild cards = [tex]C(26,3)\timesC(4,2)[/tex]

[tex]=\dfrac{26!}{3!23!}\times\dfrac{4!}{2!2!}\\\\=15600[/tex]

Probability that a hand will contain exactly two wild cards = [tex]\dfrac{15600}{142506}=0.1095[/tex]

b) Number of ways to choose two wild cards, two red cards, and one blue cards = [tex]C(4,2)\times C(10,2)\times C(5,1)[/tex]

[tex]=\dfrac{4!}{2!2!}\times\dfrac{10!}{2!8!}\times5=1350[/tex]

Probability that a hand will contain two wild cards, two red cards, and one blue cards = [tex]\dfrac{1350}{142506}=0.0095[/tex]

Answer:

The probability mass function  that you never win [tex]^xC_o[/tex] = [tex](\dfrac{999999}{1000000})^x[/tex]

Step-by-step explanation:

Given that;

the winning chance of a weekly local lottery = [tex]\dfrac{1}{10^6}[/tex]

= [tex]\dfrac{1}{1000000}[/tex]

The probability of losing = 1 - probability of winning (winning chance)

The probability of losing = [tex]1- \dfrac{1}{1000000}[/tex]

The probability of losing =[tex]\dfrac{999999}{1000000}[/tex]

The probability mass function  that you never win [tex]^xC_o[/tex] = [tex](\dfrac{1}{10^6} )^0 ( \dfrac{999999}{1000000})^x[/tex]

The probability mass function  that you never win [tex]^xC_o[/tex] = [tex](\dfrac{999999}{1000000})^x[/tex]

Step-by-step explanation:

Let minimum score on the final exam to earn an A be X

[tex]mean \: = \frac{sum \: of \: observation}{number \: of \: observation} [/tex]

[tex]90 = \frac{83 + 97 + 89 + 82 + x}{5} [/tex]

Further solving :

X = 99 marks