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Sergio039 [100]
3 years ago
5

8640 divided by40 long division

Mathematics
1 answer:
Tju [1.3M]3 years ago
7 0
The answer is 216.
https://photomath.net/s/YLLqQX
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Do you can download 20 songs in five minutes how long does it take to download one song
Eva8 [605]

Answer:

4

Step-by-step explanation:

20 divided by 5 equals 4

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3 years ago
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Find the circumference.<br> Use 3.14 for T.<br> T=pie<br> r = 2 m<br> C = [?] m<br> C= Td
vivado [14]

Answer:

314 square centimeters

Step-by-step explanation:

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2 years ago
Find the distance between xequals[Start 2 By 1 Matrix 1st Row 1st Column 8 2nd Row 1st Column negative 5 EndMatrix ]and yequals[
kherson [118]

Answer:

Distance =√(x₁ - y₁)²+ (x₂ - y₂)² = √97 = 9.85

Step-by-step explanation:

The Matrix X and Y could also be referred to as vectors in Rⁿ dimensions.

if Vector X = ( x₁ , x₂)   and  Vector  Y = (y₁ , y₂)

then, Distance (X-Y) = ||X-Y|| = √(x₁ - y₁)²+ (x₂ - y₂)²

where, x₁ = 8, x₂ = -5   and  y₁ = -1 , y₂ = -9

Distance = √(8 - (-1))²+ (-5 - (-9))² = √9² + 4² =√97 = 9.85

4 0
3 years ago
Why is it important to line up the decimal points?
jolli1 [7]
The operations must take account of the place values of individual digits in the numbers.

Please give brainliest :)))
5 0
3 years ago
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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
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