In the sequence: 13, 17, 19, 24, 27, ..., what number should come next?

mr. hapwe is buying new cowboy hats. All together the hats cost 75.12., The second hats cost twice as much as the first hat. wha

Anna11 [10]

The answer is $50.08.

6 0

4 years ago

QUICK I AM TIMED WILL MARK BRAINLIEST

Viktor [21]

I think it’s a and c might be wrong though

6 0

3 years ago

A concession stand sell hot dogs and hamburgers. At a football game, 84 hot dogs and 36 hamburgers were sold for $276. At anothe

ICE Princess25 [194]

Answer:

The concession stand sold 46 hot dogs and 32 hamburgers.

Step-by-step explanation:

The first thing to do in algebraic problems is assign variables to things we don't know, so let's start there:

We don't know how many hot dogs the concession stand sold, so we will call that number d.

We don't know how many hamburgers the concession stand sold, so we will call that number h.

Now we translate the statements into algebraic equations:

The number of hot dogs and hamburgers that were sold is 78, so d+h=78.

If each hot dog is sold for 1.25, then the total revenue from hot dogs is given by 1.25d. In the same way, the total revenue from hamburgers is 1.50h. The total revenue from both hot dogs and hamburgers should be the sum of these, and since we are told the total revenue is 105.50, we can say 1.25d+1.5h=105.5.

We now have a system of two linear equations:

d+h=78

1.25d+1.5h=105.5

We can solve it using several methods, though I'm going to go with substitution. Use the first equation to solve for d:

d+h=78

→d=78−h

Now plug this in for d in the second equation:

1.25d+1.5h=105.5

→1.25(78−h)+1.5h=105.5

Solving for h, we have:

97.5−1.25h+1.5h=105.5

0.25h=8

h=8.25→h=32

Since h+d=78,

32+d=78→d=46

The concession stand therefore sold 46 hot dogs and 32 hamburgers.

3 0

3 years ago

Which ratio is equivalent to the ratio 18:24

earnstyle [38]

Hahah shahs hahah hshshshhshs

8 0

3 years ago

Read 2 more answers

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