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3 years ago

What's the answer? Explain.

Mathematics

1 answer:

pav-90 [236]3 years ago

4 0

X = 1/3y
First step is to multiply both sides by 3 to clear the fraction.
3x = y
Next subtract y from both sides. 3x -y = 0

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Chaz is a 79.2 pound dog that was just adopted. Based on his foster home his daily energy intake is 1,840 kcal. Chaz’s new owner

zimovet [89]

Answer:

option (c)5 and $\frac{\textup{1}}{\textup{4}}$ cups

Step-by-step explanation:

Given:

Weight of Chaz = 79.2 pounds

Daily energy intake requirement = 1,840 kcal

Energy present in different brand food = 3,523 kcal/kg

Therefore,

the amount of food required = $\frac{\textup{Daily energy intake requirement}}{\textup{Energy present in different brand food}}$

the amount of food required = $\frac{\textup{1,840}}{\textup{3,523}}$

the amount of food required = 0.52228 kg

also,

1 kg = 1000 grams

therefore,

the amount of food required = 0.52228 × 1000 = 522.28 grams

now,

A cup of food weighs 100 grams

hence,

the number of cups required = $\frac{\textup{the amount of food required }}{\textup{Amount of food per cup}}$

the number of cups required = $\frac{\textup{522.28}}{\textup{100}}$

the number of cups required = 5.2228 ≈ 5 and $\frac{\textup{1}}{\textup{4}}$ cups

Hence, the correct answer is option (c)

4 0

3 years ago

Store A has a weekend sale. To the nearest hundredth of a percent, what discount must Store A offer for you to buy the system th

Ad libitum [116K]

We are given the cost of the product along with the markup percentage. The markup is how the store makes a profit. To figure out the markup price, simply multiply the "cost to store" by 100% plus the markup percent.

So the markup prices per store is:

A = 162 * 1.4 = $226.80
B = 155 * 1.3 = $201.50
C = 160 * 1.25 = $200.00

Clearly, the best deal is store C. So in order to choose store A, we want to pay less than the price of Store C, $200. So if Store A cost $199.99, then we would purchase there instead of Store C.

Therefore, $226.80 - $199.99 = $26.81
If Store A offered a discount of $26.81, we would purchase from Store A.

$26.81/$226.80 = 0.1182

So we would need a discount of 11.82% from Store A.

7 0

4 years ago

There are x ducks in a pond. The number of swans in the pond is equal to the square root of the number of ducks. If the total nu

Mashcka [7]

Answer:

<h2>There are 9 ducks and 3 swans in the pond.</h2>

Step-by-step explanation:

According to the problem, the number of swans in the pond is equal to the square root of the number of ducks, this can be expressed as

$s=\sqrt{x}$

Where $x$ represents the number of ducks in the pond.

If the total number of swans and ducks is 12, the expression would be

$s+x=12$

Now, we replace the first equation in the second one

$\sqrt{x} +x=12$

Then, we elevate each side to the square power

$\sqrt{x} =12-x\\(\sqrt{x} )^{2} =(12-x)^{2} \\x=144-24x+x^{2} \\x^{2} -25x+144=0$

Using a calculator, the roots are 9 and 16.

Let's evaluate each value.

For $x=9$ , $s=\sqrt{9}=3$ , and $9+3=12$ , which is true.

For $x=16$ , $s=\sqrt{16} =4$ , and $16+4=20$ , which is false.

Therefore, there are 9 ducks and 3 swans in the pond.

7 0

3 years ago

Read 2 more answers

Urgen helpppppp please

GrogVix [38]

I think the correct answer iS D

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