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2 years ago

Please answer the question from the attachment.

Mathematics

1 answer:

pochemuha2 years ago

3 0

35 degrees Celsius because 5÷9 is .333 and .333×(95-32)= About 35. Hope that helped :D

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If (n+1)!=90×(n-1)!,find n.

Helen [10]

$\displaystyle\bf\\(n+1)!=90\times (n-1)!\\\\(n-1)!\times n\times(n+1)=90\times (n-1)!~~~\Big|~:(n-1)!\\\\n(n+1)=90\\\\n(n+1)=9\times10\\\\n(n+1)=9(9+1)\\\\\implies~~\boxed{\bf n=9}$

4 0

2 years ago

Find c.<br> Round to the nearest tenth.

solmaris [256]

Answer: Picture is blury for me I can not see it well enough

Step-by-step explanation:

All i see is the 8 ft and 17ft cant see any of the other numbers

8 0

2 years ago

A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic

aliya0001 [1]

Answer:

P-value = 0.4846

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4.73

Sample mean, $\bar{x}$ = 4.35

Sample size, n = 51

Alpha, α = 0.05

Sample standard deviation, s = 3.88

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 4.73\\H_A: \mu \neq 4.73$

We use Two-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{4.35 - 4.73}{\frac{3.88}{\sqrt{51}} } = -0.699$

Now, $z_{critical} \text{ at 0.05 level of significance } = \pm 1.96$

We can calculate the p-value with the help of standard normal table.

P-value = 0.4846

Since the p-value is higher than the significance level, we fail to reject the null hypothesis and accept it.

We conclude that this college has same drinking habit as the college students in general.

7 0

3 years ago

What is the answer to this question?

Nitella [24]

Answer:

11x+2

Step-by-step explanation:

4 0

2 years ago

Preform the indicated operation.<br> 4/11 ÷ 4/9

BartSMP [9]

Answer:

0/8

Step-by-step explanation: i tried my best

4 0

2 years ago

Read 2 more answers