All categories

3 years ago

Which problem matches the equation | 75 ≤ 16x + 10

Mathematics

2 answers:

seropon [69]3 years ago

8 0

Answer:

work on each problem when assigned, since the questions you may have realize the distinct methods and philosophies of teaching at Exeter. in 6 hours working alone, how long would it take Kit to build the wall when working alone? Using a graphing tool, with the window set as −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10,.

Step-by-step explanation:

beks73 [17]3 years ago

7 0

Answer:

B - Jesse practices his guitar for less than 75 hours a month. He practices independently 16 times for x hours and has 10 lessons a month. How many hours could Jesse practice independently each time?

Step-by-step explanation:

Correct me if I'm wrong!!

You might be interested in

Complete the statement with the choice that best

chubhunter [2.5K]

Answer:

Factors of the constant term

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Find the unit price for each option shown below. Round to the nearest cent when necessary. Option I: 10 candy bars for $6.75 Opt

natita [175]

Option I: 10 candy bars for $6.75

6.75/10 = 0.68 

Option II: 12 candy bars for $7.25 

7.25 / 12 = 0.60

6 0

3 years ago

Read 2 more answers

Which answer is a graph for the inequality x>2

seropon [69]

Show me the answers, then i'll help

6 0

3 years ago

The sum of the digits of a certain two-digit number is 7. Reversing its digits increases the number by 9. What is the number? wr

dimaraw [331]

Xy is the number

A) x + y = 7
B) 10y + x = 10x + y + 9
collecting terms of equation B
B) -9x + 9y = 9
multiplying A) by 9
A) 9x + 9y = 63 then adding it to equation B)
B) -9x + 9y = 9
18y = 72
y = 4
x = 3

The number is 34

4 0

3 years ago

After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni

antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

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

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.64=64>10$

$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

6 0

3 years ago