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svetoff [14.1K]

3 years ago

9

13. The altitude of an equilateral triangle is 6 in. long. Find the length of a side of the triangle. (Hint: what triangles are

formed by the altitude?)

1 answer:

JulijaS [17]3 years ago

6 0

Yeah the answer is in that file

You might be interested in

62=−3k+7(k+6)+4<br><br> solve for k

guajiro [1.7K]

Step-by-step explanation:

62= -3k + 7k + 42 + 4

62 - 42 - 4 = 4k

16 = 4k

4 = k

7 0

3 years ago

madison middle school spends$750 per year on club activities.they soend at least twice that amount on after -school activities.w

aev [14]

X≥1500
x represents how much money is spent on after school activities

8 0

4 years ago

During the 2000 season, the home team won 138 out of 240 regular season National Football League games. (15 points) a) Construct

jeka94

Answer:

Step-by-step explanation:

A) Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 240

x = 138

p = 138/240 = 0.58

q = 1 - 0.58 = 0.42

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, the z score for a confidence level of 95% is 1.96

Therefore, the 95% confidence interval is

0.58 ± 1.96√(0.58)(0.42)/240

Confidence interval is

0.58 ± 0.062

B) winning more than halve of the games would be winning 120 games and above.

p = 120/240 = 0.5

We would set up the hypothesis test.

For the null hypothesis,

P ≥ 0.5

For the alternative hypothesis,

P < 0.5

Considering the population proportion, probability of success, p = 0.5

q = probability of failure = 1 - p

q = 1 - 0.5 = 0.5

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 138

n = number of samples = 240

P = 138/240 = 0.58

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.58 - 0.5)/√(0.5 × 0.5)/240 = 2.48

Recall that this is a left tailed test. We would determine the probability value of the area to the right of the z score from the normal distribution table.

P value = 1 - 0.9934 = 0.0066

Since alpha, 0.01 > the p value, 0.0066, then we would reject the null hypothesis.

Therefore, At the 0.01 significance level, there is no strong evidence of a home field advantage (they win more than half of the games) in professional football.

3 0

3 years ago

The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by World One Research, included the question, "H

LenaWriter [7]

Answer:

A 95% confidence interval estimate of the mean number of hours a legal professional works on a typical workday is [7.44 hours, 10.56 hours].

Step-by-step explanation:

We are given that x is normally distributed with a known standard deviation of 12.6.

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

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

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average mean response = 9 hours

$\sigma$ = population standard deviation = 12.6

n = sample of legal professionals = 250

$\mu$ = mean number of hours a legal professional works

<em>Here for constructing a 95% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $9-1.96 \times {\frac{12.6}{\sqrt{250} } }$ , $9+1.96 \times {\frac{12.6}{\sqrt{250} } }$ ]

= [7.44 hours, 10.56 hours]

Therefore, a 95% confidence interval estimate of the mean number of hours a legal professional works on a typical workday is [7.44 hours, 10.56 hours].

5 0

3 years ago

Six groups of students sell 162 balloons at the school carnival. there are 3 students in each group. if each student sells the s

Marina CMI [18]

Since there are six groups of student and each group consists of 3 students, therefore, the total number of students is 18. All of them sold 162 balloons. In order to know how many balloons each student sold, we just have to divide 162 by 18 and the answer is 9. Hope this answers your question.

4 0

3 years ago