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Ksenya-84 [330]

2 years ago

12

If using the method of completing the square to solve the quadratic equation x^2+8x+37=0, which number would have to be added to

"complete the square"?

1 answer:

Feliz [49]2 years ago

6 0

Answer:

The numbers are --> x = -4 + √21 i and x = -4 — √21 i

See the attachment for my step-by-step explanation

(solution is in red box)

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Penny's garden has alength of 16ft. and the width is 5 ft. with a rectangular fountain in the middle that measures 5 ft by 9 ft.

lilavasa [31]

First we need the whole garden:

16 x 5 = 80

Now we need to fountain:

5 x 9 = 45

Now to find our answer:

80 - 45 = 35

So our answer is B. 35ft²

4 0

2 years ago

A square garden has an area of 1,600 square feet. How long is each side of

kiruha [24]

Answer:

40

Step-by-step explanation:

why? because 40x40=1600

6 0

2 years ago

hanna buys 3 packs of 36 exposure film and 2 packs of 24 exposure film. she uses 8 rolls of film. how many rolls does she have l

RSB [31]

To solve this problem, first we have to figure out how many total rolls Hanna buys.

3 packs*36 rolls in each pack = 108 total rolls
2 packs*24 rolls in each pack = 48 total rolls

108 rolls + 48 rolls = 156 rolls.

Next, it says that Hanna uses 8 rolls of film. To model this in our expression, we need to subtract 8 rolls from our total amount.

156-8 = 146 rolls

Hanna has 146 rolls left.

7 0

2 years ago

Read 2 more answers

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The follow

kipiarov [429]

Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

$H_0: \mu = 13252$

At the alternative hypothesis, we test if the mean cost has increased, that is:

$H_1: \mu > 13252$

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

13252 is tested at the null hypothesis:

This means that $\mu = 13252$

The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

This means that $n = 20, X = 15560, s = 3500$

Value of the test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}$

$t = 2.95$

P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

7 0

2 years ago

preliminary sample of 100 labourers was selected from a population of 5000 labourers by simple random sampling. It was found tha

VladimirAG [237]

Answer:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$

n=369

Step-by-step explanation:

1) Notation and definitions

$X=40$ number of the selected labourers opt for a new incentive scheme.

$n=100$ random sample taken

$\hat p=\frac{40}{100}=0.4$ estimated proportion of the selected labourers opt for a new incentive scheme.

$p$ true population proportion of the selected labourers opt for a new incentive scheme.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Solution tot he problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$

And rounded up we have that n=369

8 0

2 years ago