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3 years ago

Find the area of the isosceles triangle

Mathematics

2 answers:

AveGali [126]3 years ago

4 0

175m^2
I got the by multiplying 1/2, by the base (14), by the height (25).

1/2 * 14 * 25

icang [17]3 years ago

4 0

First we are gonna find the height using trigonometry.

h^2 = 25^2 - 7^2
h = 24

Height is 24m

Now find area using 1/2 x b x h

14/2 x 24 = 168 m^2

Therefore area is 168m^2

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Find the volume of the composite solid. Round your answer to the nearest hundredth. A. 22.05mm B. 22.19mm C. 22.53mm D. 22.54mm

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3 years ago

(Lesson 31) The population of Jacksonville, Florida, was 736,000 in 2000 and was increasing at a rate of 1.49% each year. At

damaskus [11]

Answer:

Step-by-step explanation:

Population t years after 2000 = 736,000 × 1.0149^t

736,000 × 1.0149^t = 1,000,000

1.0149^t = 1,000,000/736,000 = 125/92

Take log base 1.0149 of both sides

t ≈ 20.7

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The table shows the numbers of students at a high school who take Spanish or Chinese. No students take both languages. What

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Step-by-step explanation:

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3 years ago

when sunayana purcased a video-game she paid Rs 162.50 as VAT at the rate of 13%. find its cost without levying VAT. Rs 1250 is

horsena [70]

Answer: 1250

Step-by-step explanation:

Let x be the cost before vat

13/100x = 162.50

x = 162.50 x 100/13

x = 162.50 / 2.13

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4 years ago

Sample Size for Proportion As a manufacturer of golf equipment, the Spalding Corporation wants to estimate the proportion of gol

Dima020 [189]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56$

And rounded up we have that n=2663

Step-by-step explanation:

Previous concepts

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

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

Solution to the 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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

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 $ME =\pm 0.025$ 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)

We can assume an estimated proportion of $\hat p =0.5$ since we don't have prior info provided. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56$

And rounded up we have that n=2663

6 0

3 years ago