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3 years ago

Please help me out! TY :D

Mathematics

2 answers:

solong [7]3 years ago

6 0

Answer:

yes google has moat of the answers for these

AlladinOne [14]3 years ago

5 0

Answer: I’m sorry if I didn’t answer to this question, but you should try askin a friend or teacher. But the best is google

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I need this ASAP but it’s ok if you see it late

klasskru [66]

Answer:

72 square units

Step-by-step explanation:

move above the dotted line to make a rectangel then the numbers are 6 and 12 then times them to get 72

7 0

2 years ago

Thirty car pass through a certain toll booth each minute.at that rate,how many cars are expected to pass through the toll booth

BigorU [14]

D. 30 Cars p/m x 30 min = 900

8 0

3 years ago

Read 2 more answers

I NEED HELP ASAP! THANKS :)

4vir4ik [10]

Answer:

Last option: increasing: -2 < x < 0 and x> 3/2;

decreasing: x < -2 and 0 < x < 3/2

Step-by-step explanation:

The function given is an absolute cubic function. (Pictured below).

We can determine from the graph that the equation increases over the interval:

increasing: -2 < x < 0 and x> 3/2;

decreasing: x < -2 and 0 < x < 3/2

6 0

3 years ago

Read 2 more answers

Plz help ASAP the number that is cut off on 69 is 24

Minchanka [31]

Answer:

$A=256\ units^2$

Step-by-step explanation:

The given figure shows a triangular pyramid having 16 units as height. The base of pyramid is triangle.

The area of a triangular pyramid is given by :

$A=\dfrac{1}{3}bH$

Where

b is the area of base

$A=\dfrac{1}{3}\times \dfrac{1}{2}\times B\times h\times H\\\\=\dfrac{1}{6}\times 12\times 8\times 16\\\\=256\ units^2$

So, the required area is equal to $256\ units^2$ .

7 0

3 years ago

A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the

Lubov Fominskaja [6]

Answer:

The required sample size for the new study is 801.

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

25% of all adults had used the Internet for such a purpose

This means that $\pi = 0.25$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

What is the required sample size for the new study?

This is n for which M = 0.03. So

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

$0.03 = 1.96\sqrt{\frac{0.25*0.75)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2$

$n = 800.3$

Rounding up:

The required sample size for the new study is 801.

4 0

2 years ago