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3 years ago

How do I solve 4(5(6)) This is a hypothetical question

2 answers:

6 0

PEMDAS states that the value in the parenthesis should be evaluated first:
4(5*6)
4(30)
After just multiply the value in the parenthesis by the number outside of the parenthesis:
120

Flura [38]3 years ago

4 0

Im not to sure but would it be 120?

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Find the tangent of angle Θ in the triangle below.

Rina8888 [55]

Answer: $\frac{\sqrt{65}}{4}$

Step-by-step explanation:

By the Pythagorean theorem, the unknown side has length

$\sqrt{9^{2}-4^{2}}=\sqrt{65}$

Therefore,

$\tan \theta=\boxed{\frac{\sqrt{65}}{4}}$

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2 years ago

You and two friends share 7 cookies equally. How many cookies do you each get?

Ivahew [28]

4 cookies because three divided by seven equals three hope I helped

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3 years ago

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If a polygon is a square, then it is also a rectangle. True or False

Art [367]

Answer:

true

Step-by-step explanation:

a square will always be a rectangle but a rectangle will not always be a square.

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3 years ago

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Factor: (m+2)^2−(m+2)+mn·(n+2)+2n(n+2)

bazaltina [42]

ANSWER
${(m + 2)}( {n}^{2} + 2n + m + 1)$

EXPLANATION

The given expression is

${(m + 2)}^{2} - (m + 2) + mn(n + 2) + 2n(n + 2)$

We factor to get,

$= {(m + 2)}(m + 2 - 1) +n (n + 2) (m+ 2)$

This implies that,

$= {(m + 2)}(m + 1) +n (n + 2) (m+ 2)$

We factor further to obtain,

$= {(m + 2)}((m + 1) +n (n + 2))$

or

$= {(m + 2)}( {n}^{2} + 2n + m + 1)$

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3 years ago

Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i

OLEGan [10]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

$H_0:\mu=18\\\\H_a:\mu\neq18$

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size : n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : $\overline{x}=18.8$

Standard deviation : $\sigma=2.8$

Test statistic for population mean :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02$

The p-value= $2P(z>2.02)=0.0433834$

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

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3 years ago