All categories

3 years ago

I need help with These

Mathematics

1 answer:

Kamila [148]3 years ago

6 0

I'm not sure about d.

This was what I could come up with.

You might be interested in

Quadrilateral RSTU is a parallelogram. What must be the value of x?

Luba_88 [7]

I set x-3 and 3x-13 equal to each other and then got x equal to 5. So, I believe the correct answer would be C.) 5

7 0

3 years ago

Read 2 more answers

Ninety tickets were sold by our school for the band fest. a tickets were sold for $3.00 each and b tickets were sold for $4.00 e

Lesechka [4]

Answer: 60= $3.00 and 30= $4.00

x=# of $3 tickets

90-x=# of $4 tickets

$3x+$4(90-x)=$300

3x+360-4x=300

-x+360=300

-x=-360+300

-x=-60

x=60 $3 tickets

90-60=30 $4 tickets

5 0

2 years ago

Find the square roots.

ahrayia [7]

Step-by-step explanation:

I think the answer would be

=1/31 and 1/31

6 0

3 years ago

Read 2 more answers

Given the measures a = 10, b = 40, and A = 30°, how many triangles can possibly be formed?

arlik [135]

Answer:

no triangle can be formed with the given values (i.e 0 triangle).

Step-by-step explanation:

Given;

length of side a, = 10

length of side b, = 40

angle A, = 30°

Apply sine rule to determine the angle of the second side (B);

$\frac{sin \ A}{a} = \frac{sin \ B}{b} \\\\\frac{sin \ 30^0}{10} = \frac{sin \ B}{40}\\\\sin \ B = 40(\frac{0.5}{10} )\\\\sin \ B = 2$

The maximum sine function is 1, there is no sine function that will be equal to 2, so there is no triangle that can be formed with the given values.

8 0

2 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