Solve d/14-1/2=1/4 Show your work.

What is the missing length???

Kaylis [27]

Answer:

25

Step-by-step explanation:

144/60 = 2.4

60/2.4 = x

x = 25

8 0

3 years ago

The diagram shows a circle inside a square 16cm Work out the area of the circle

vesna_86 [32]

i can help if you have the picture

5 0

3 years ago

Read 2 more answers

Serena estimates that she can paint 60 square feet of wall space every half-hour. Write an equation for the relationship with ti

Katarina [22]

Answer:

So, the relationship with i hours as the indecent variables would be

Serena can't paint 400 square feet of wall space of wall space in 3.5 hours.

Step-by-step explanation:

Since we have given that

Area that she can paint = 60 square feet

Time taken = half hour = 30 minutes = 0.5 hours

Using "Unitary method, we get that

In 0.5 hours, she can paint = 60 square feet

In 1 hour, she can paint =

In i hours, she can paint =

Let the total area she can paint be 'y'.

So, the relationship with i hours as the indecent variables would be

Now, Can Serena paint 400 square feet of wall space i. 3.5 hours

We put y = 400 and i = 3.5 hours to check whether it is equal or not.

Hence, Serena can't paint 400 square feet of wall space of wall space in 3.5 hours.

6 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

HELP HELP HELP HELP HELP

Simora [160]

Answer : 32
( angles of a triangle add up to 180)

6 0

3 years ago