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andriy [413]
2 years ago
10

The area of a rectangular patio is 24 3/8

Mathematics
1 answer:
scoray [572]2 years ago
5 0

Answer:

6 1/2

Step-by-step explanation:

Use the area formula: A=LengthxWidth

Plug in what we know: 24 3/8= 3 3/4x Width

We need to isolate the width; get it by itself.

Do this by (24 3/8)/(3 3/4). The best way is to enter this into a calculator.

Your answer is 6 1/2

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6 kg of potatoes cost $12.90. How many kilograms of potatoes can you get with $8? Round to the nearest whole number.
Mandarinka [93]

Answer:

4 kg

Step-by-step explanation:

Answered By Huntermike976  

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4 0
3 years ago
Solve for x if 10/4x-3 = 2​
Marysya12 [62]

Answer:

We need to solve this equation:

\frac{10}{4x}-3=2

We pass -3 to the right summing:

\frac{10}{4x} = 2 + 3

\frac{10}{4x}=5

We pass the 4x that is dividing to the right multiplying:

10 = 5*4x

10=20x

We pass the 20 of the right to the left dividing:

\frac{10}{20}=x

x=\frac{1}{2}

Now, we know x=\frac{1}{2}, so we can verify:

\frac{10}{4*\frac{1}{2}}-3 = 2

\frac{10}{2}-3=2

5 - 3 = 2 TRUE

3 0
3 years ago
Read 2 more answers
Use the information to determine the exact trigonometric value sin theta =-1/5, pi
butalik [34]

Answer: -2 (sqrt:6)/5

Step-by-step explanation:

5 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
Who knows how to resolve this
Colt1911 [192]
You have to distribute the radical negative 18 in first. Simplifying from there, you’ll prob have to get into imaginary numbers since the square root of a negative is imaginary
8 0
3 years ago
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