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3 years ago

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:

x = 6 over 5

Step-by-step explanation:

mark me brainliest if u have any doubt ask me

4 0

3 years ago

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

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

Pr(8) = $\frac{1}{25}$

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

Multiples of 5 = 5, 10, 15 and 25

Pr (multiples of 5) = $\frac{4}{25}$

Pr of drawing a card with odd numbers:

Number of odd numbers between 1 and 25 = 13

Pr(odd numbers) = $\frac{13}{25}$

Pr of drawing a number with square number on it;

Square numbers between 1 and 25 = 1, 4, 9, 16 and 25

Pr(square numbers) = $\frac{5}{25}$ = $\frac{1}{5}$

3 0

3 years ago

Great deal of work to do. Which word is the closest meaning to the underlined word

GarryVolchara [31]

What is the underlined word ??

3 0

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

Step-by-step explanation:

base = 6

area = 15

area = 1/2(6)h

15= 3h

5=h

4 0

3 years ago