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2 years ago

Find the coordinates of the midpoint of AL¯¯¯¯¯¯¯ with endpoints A(–4, 3) and L(6, 7).

Mathematics

1 answer:

iogann1982 [59]2 years ago

6 0

thanks for point 40_56yghuuc

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QUICK WHAT IS:which expression represents a factorization of 32m + 56mp

Anna71 [15]

Simplifying that expression will be
32m + 56mp = 8m (4 + 7p)

5 0

3 years ago

Classify the following triangle

larisa86 [58]

Answer:

A. Right Triangle

D. Isosceles

Step-by-step explanation:

Well, since it has a square at the bottom of the triangle, it means it's a right triangle, because it has a 90 degree angle.

It is also an isosceles triangle, because two of the sides are of equal length.

Hope this helped!

4 0

3 years ago

Of video game that originally cost $25.99 is on sale for 50% off if you have $14 would you have enough money to buy the video ga

NeX [460]

Just round 25.99 to 26 and divide it by two. This gives you 13, which is less than 14.. You would be able to buy the video game.

3 0

3 years ago

How many 6-digit numbers can be created using8, 0, 1, 3, 7, and 5 if each number is used only once?

Ludmilka [50]

Answer:

600 numbers

Step-by-step explanation:

For six-digit numbers, we need to use all digits 8,0,1,3,7,5 each once.

However, 0 cannot be used as the first digit, because it would make a 5-digit number.

Therefore

there are 5 choices for the first digit (exclude 0)

there are 5 choices for the first digit (include 0)

there are 4 choices for the first digit

there are 3 choices for the first digit

there are 2 choices for the first digit

there are 1 choices for the first digit

for a total of 5*5*4*3*2*1 = 600 numbers

6 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

2 years ago