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2 years ago

10

What is this problem

1 answer:

sergeinik [125]2 years ago

3 0

Answer:

$m=A-\frac{2n}{Z}$

Step-by-step explanation:

Multiply by 2:

2n = Z(A - m)

Divide by Z:

$\frac{2n}{Z} =A-m$

Subtract A:

$-m=\frac{2n}{Z}-A$

Multiply by -1:

$m=A-\frac{2n}{Z}$

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Someone please help me with question 2

alukav5142 [94]

Answer:

v = s/t

v = (381.59 miles) / (5.5 hours)

v = 69.38 mph

8 0

2 years ago

Pls help me divide this radical and pls show work

Alenkinab [10]

Let's start by rationalizing the radical:

4√6/√50 * √50/√50

This gets rid of the radical in the denominator, giving us:

4√300/50

√300 can be simplified to √3 * √100 or 10√3. Now we can insert this back into the numerator:

4 * 10√3/50

This is equivalent to:

40√3/50

40/50 can be simplified to 4/5.

Therefore, the answer is 4√3/5.

5 0

2 years ago

rancho middle school has an average of 23/3/8 students per teacher. write this mixed number as a decimal

Len [333]

Rancho middle school has an average of 23 3/8 students per teacher.write this mixed number as a decimal.

Answer:

$23\frac{3}{8} = 23.375$

Step-by-step explanation:

Given:

The given mixed number is $23\frac{3}{8}$

We need to write the given mixed number as a decimal.

So, we find the decimal number by separating the parts.

$=23+\frac{3}{8}$

The value of $\frac{3}{8}=0.375$

$=23+0.375$

$=23.375$

Therefore, the decimal number of given mixed number is $=23.375$

8 0

2 years ago

Wayne plays multiple instruments in a rock band. He also composes the songs. He takes 36 weeks to write the songs for an album o

sammy [17]

He took 36 weeks to compose 12 songs.

Unit rate : Divide total weeks by total songs:

Unit Rate = 36 / 12 = 3 weeks per song.

5 0

3 years ago

Read 2 more answers

Preliminary data analyses indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the

Readme [11.4K]

<h2>Answer with explanation:</h2>

By considering the given information, we have

Null hypothesis : $H_0: \mu=35.0$

Alternative hypothesis : $H_1: \mu\neq35.0$

Since the alternative hypothesis is two-tailed , so the test is a two-tailed test.

Given : Sample size : n= 20, since sample size is less than 30 so the test applied is a t-test.

$\overline{x}=41.0$ ; $\sigma= 3.7$

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

i.e. $t=\dfrac{41.0-35.0}{\dfrac{3.7}{\sqrt{20}}}=7.252112359\approx7.25$

Degree of freedom : n-1 = 20-1=19

Significance level = 0.01

For two tailed, Significance level $=\dfrac{0.01}{2}=0.005$

By using the t-distribution table, the critical value of t = $t_{19, 0.005}=2.861$

Since , the observed t-value (7.25) is greater than the critical value (2.861) .

So we reject the null hypothesis, it means we have enough evidence to support the alternative hypothesis.

We conclude that there is some significance difference between the mean score for sober women and 35.0.

3 0

2 years ago