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statuscvo [17]
3 years ago
7

Need help with number 42 plz. Math.

Mathematics
1 answer:
Trava [24]3 years ago
6 0

so this is like cancelling out. so if you have positive 13 and have an answer of 3 how much do you need to subtract/ cancel out to get there? 10

I think you can get the rest


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X over 3 equals 2 over 5
natali 33 [55]

Answer:

so

x = 6 over 5

Step-by-step explanation:

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Please help emergency
Firdavs [7]

Step-by-step explanation:

This is a probability related problem.

  Probability is the likelihood of an event to occur;

    Pr = \frac{number of possible outcomes}{Total sample space}

The sample space here is from 1 to 25 which is 25

A.

Pr of a card marked 8; we have just 1 possible outcome;

       Pr(8)  = \frac{1}{25}

B.

Pr  of drawing a card that is a multiple of 5;

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C.

Pr of drawing a card with odd numbers:

          Number of odd numbers between 1 and 25  = 13

     Pr(odd numbers) = \frac{13}{25}

D.

Pr of drawing a number with square number on it;

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3 years ago
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GarryVolchara [31]
What is the underlined word ??
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3 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
Find the height of a triangle if the base measures 6 inches and the area is 15 square inches. Use the formula for the area of a
harkovskaia [24]

Answer:

height equals 5 inches

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base = 6

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area = 1/2(6)h

15= 3h

5=h

4 0
3 years ago
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