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2 years ago

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a triangle has sides of 4 and 6 units. The measurement of the longest side is missing.Ted says that one possibility for the unkn

own side length is 11, Do you agree with ted

1 answer:

tester [92]2 years ago

7 0

This is called the pythagorean theorem: a^2 + b^2 = c^2.
Basically, the sum of one side squared + the side of another side squared = the length of the longest side (or hypotenuse).

We have to do the following math:
4 * 4 = 16
6 * 6 = 36
16 + 36 = 52.

We know that 52 = c^2
So we have to $\sqrt{c}$ to get 7.21

You do not agree with Ted.

You might be interested in

Assume that a fair die is rolled. The sample space is , 1, 2, 3, 4, 56, and all the outcomes are equally likely. Find P8. Expres

liraira [26]

Answer:

$P\left(8\right)=0$

Step-by-step explanation:

Assuming that a fair die is rolled.

The sample space is 1, 2, 3, 4, 5, 6 and all the outcomes are equally likely.

Let X be the set of all possible outcomes. Let A be an outcome.

So, the probability that A occurs is:

$P\left(A\right)=\frac{|A|}{|X|}$

As the set of all possible outcomes of the roll of a single die is:

$X=\left\{1,2,3,4,5,6\right\}$

Observe that

$|X|=6$

Here

$|A|=0$ because 8 is not in the set sample space. So, the outcome of occurring the number 8 is not possible from the all possible outcomes.

So, the probability must be zero.

In other words,

$P\left(A\right)=\frac{|A|}{|X|}$

$P\left(8\right)=\frac{0}{6}=0$

Therefore,

$P\left(8\right)=0$

5 0

2 years ago

For the lunch special at a high school cafeteria, students can get either salad or french fries as a side order. The following t

LekaFEV [45]

Answer: E

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

The national average for mathematics SATs in 2011 was 514 and the standard deviation was approximately 40. a) Within what bounda

blondinia [14]

Answer:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

$\mu = 514 , \sigma =40$

For this case we can use the Chebysev theorem who states that "At least $1 -\frac{1}{k^2}$ of the values lies between $\mu -k\sigma$ and $\mu +k\sigma$ "

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

4 0

2 years ago

The prices of three t-shirts styles are $24, $30 and $36. the probability of choosing a $24 t-shirt is 1/6. the probability of c

Slav-nsk [51]

$\text{Answer} : \text{The expected value of a t-shirt is \$31.}$

Explanation:

Since we have given that

The prices of three t-shirts styles i.e $24, $30, $36 with their probability is given by

$\frac{1}{6}, \frac{1}{2},\frac{1}{3}$

As we know that,

$E(X)= \sum_{1}^{3}x_iP(x_i)$

$\text{where} x_i \text{ is the prices of t- shirts styles}$

Now,

$x_1= \$24 , x_2=\$30 , x_3=$36$

and

$P(x_1)=\frac{1}{6},P(x_2)=\frac{1}{2}, P(x_3)=\frac{1}{3}$

So,

$E(X)= 24\times \frac{1}{6}+30\times\frac{1}{2}+36\times \frac{1}{3}\\=4+15+12\\=31$

So, the expected value of a t-shirt = $31.

4 0

2 years ago

a university theater sold 556 tickets for a play, tickets cost $22 per adult tickets and $ 12 for senior citizen ticket, if the

Deffense [45]

Let no. of adult tickets be x and senior citizen tickets be y. x + y = 556 22x + 12y = 8492 Solving by simultanous equations, x = 182 adult tickets y = 374 senior citizen tickets

7 0

2 years ago