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3 years ago

I need help with my algebra 1 extra credit anything helps

Mathematics

1 answer:

Crazy boy [7]3 years ago

8 0

You dont have a question

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HELP PLEASE DUE IN 3 MINUTES

Morgarella [4.7K]

Answer: it is d becaues that is the mode

4 0

3 years ago

In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected fo

larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine $z_{\frac{\alpha}{2}}=z_{0.005}$

Now by using z score table we find that $z_{\frac{\alpha}{2}}=2.58$

The boundaries of the confidence interval are:

$\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64$

Hence, the 99% confidence interval for the population mean is 22.96 to 26.64

3 0

4 years ago

Lesson 10 adding and subtracting in scientific notation

Gnoma [55]

What about lesson 9?

3 0

4 years ago

3. Amy raises $58.75 to participate in a walk-a-thon.

Leviafan [203]

Oscar raises 246$ because 58.75 + 23.25 is82 82 x 3 is 246

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3 years ago

Find the lowest common denominator of 3/x^3y and 7/xy^4

muminat

Answer:

i dont know get good at math

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers