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3 years ago

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

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Answer:

Step-by-step explanation:

20 divided by 5 equals 4

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3 years ago

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Find the circumference. Use 3.14 for T. T=pie r = 2 m C = [?] m 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

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3 years ago

Why is it important to line up the decimal points?

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The operations must take account of the place values of individual digits in the numbers.

Please give brainliest :)))

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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

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

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3 years ago