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Angelina_Jolie [31]
4 years ago
10

1. Solve the inequality. -12 < 2x - 4 <10

Mathematics
2 answers:
noname [10]4 years ago
6 0

Answer:

8<x<7

Step-by-step explanation:

12<2x−4<10

12+4<2x−4+4<10+4(Add 4 to all parts)

16<2x<14

16

2

<

2x

2

<

14

2

(Divide all parts by 2)

malfutka [58]4 years ago
4 0

Answer:

-4  < x  < 7

Step-by-step explanation:

-12 < 2x - 4 <10   (add 4 across the inequality)

-12 + 4 < 2x  <10 + 4

-8 < 2x  < 14  (divide ineqality by 2)

-8/2  < x  < 14/2

-4  < x  < 7

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"the distribution of means is the correct comparison distribution when"
alexandr402 [8]

The distribution of means is the correct comparison distribution when

Answer: The distribution of means is the correct comparison distribution when there is more than one person in a sample.

The distribution of the sample means is also called the sampling distribution of mean. It is most appropriate when we take a random sample of size n from the population of size N. The distribution of the sample means will follow normal distribution with mean = \mu and standard deviation = \frac{\sigma}{\sqrt{n}}


4 0
4 years ago
A bag holds 5 pounds of pet food. If Paul uses the 5 pounds of food to fill 6 plastic containers equally how much pet food will
serg [7]
Answer: 13 1/3 ounces

Explanation:
1 pound equals 16 ounces
5 x 16=80
80 ounces divided by 6 containers is equal to 13 1/3 ounces of pet food.
5 0
3 years ago
Mark put his dog on a diet. The dog's total weight change for the first two weeks was -3/4 pound. The dog lost the same amount o
saw5 [17]
The change in the first week is .375 pounds =. Divide 3/4 by 2
8 0
4 years ago
The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
5 0
3 years ago
If a 9-foot tall adult elephant casts an 18-foot shadow, how long is the shadow of the 6-foot zookeeper? use a proportion to set
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9 6
— x —
18 y


y is equal to 12. I’m not positive about how the work is shown though.
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4 years ago
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