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3 years ago

C3 if c = 2 need help

Mathematics

1 answer:

7nadin3 [17]3 years ago

4 0

Answer:

Step-by-step explanation:

If c is 2 then that means you would multiply 2 and 3 to get 6

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What is the value of X? Please help!!!!

Tom [10]

The value of x would be 45 because it is an equivalent triangle so the opposite side would be the same as the other side. ^-^ hope this helps.

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3 years ago

What is the value for x? Enter your answer in the box. x = ____

vlada-n [284]

180 - 73-73 =34

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3 years ago

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The

Tems11 [23]

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that $n = 387, \pi = \frac{298}{387} = 0.77$

84% confidence level

So $\alpha = 0.16$ , z is the value of Z that has a p-value of $1 - \frac{0.16}{2} = 0.92$ , so $Z = 1.405$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8$

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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3 years ago

For what values of b will F(x) = logy x be a decreasing function?

UNO [17]

Answer:

c. 0<b<1

Step-by-step explanation:

F(x) = logb x

When b is less than 1 see attached graph

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3 years ago

Read 2 more answers

G E O M E T R Y (will reward brainliest! please help)

Dennis_Churaev [7]

Is this a question or a sign up?

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4 years ago