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DENIUS [597]
3 years ago
5

Could Someone give me a run down of how these problems are done? i used a calculator to get the answers but now i want to know h

ow to do these on my own

Mathematics
1 answer:
morpeh [17]3 years ago
5 0
It’s just the same as if there were no decimal point. but if there isn’t a number in the quotient on the left of the decimal like #15, then you just place a 0 and continue to the next dividend which is 50.

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Answer:

the answer is 4

Step-by-step explanation:

the black shape times 4 is the blue shape

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Shirley is decorating the gym for a party, she has 4 rolls of blue streamer and 7 rolls of green streamer. if each roll of blue
Radda [10]
You do 4 times 95 and you get 380 plus 7 times 84 and that's 588 and 380+588=<span>968 feet.</span>
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3 years ago
6. Which of the following is the solution set to the equation x² + 8x -15 = 5x +13?
Verizon [17]

Answer:

The answer is A.......

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3 years ago
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

4 0
3 years ago
Lucy is using a one-sample t ‑test based on a simple random sample of size n = 22 to test the null hypothesis H 0 : μ = 16.000 c
slega [8]

Answer:

The value of test statistic is 1.338

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 16.000

Sample mean, \bar{x} = 16.218

Sample size, n = 22

Alpha, α = 0.05

Sample standard deviation, s = 0.764

First, we design the null and the alternate hypothesis

H_{0}: \mu = 16.000\text{ cm}\\H_A: \mu < 16.000\text{ cm}

We use one-tailed t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

t_{stat} = \displaystyle\frac{16.218 - 16.000}{\frac{0.764}{\sqrt{22}} } = 1.338

Thus, the value of test statistic is 1.338

4 0
4 years ago
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