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2 years ago

Mia got $80 for her birthday and spent $60. What percent of her birthday money does she have left

Mathematics

1 answer:

max2010maxim [7]2 years ago

4 0

Answer:

16 percent is left

Step-by-step explanation:

20/100×80

Hope it helps!!

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What is the sum of Negative 8 + 6?

Pavlova-9 [17]

Answer:

-14 or -2

Step-by-step explanation:

-8+-6= -14

-8+6 = -2

7 0

3 years ago

Sam wants to fence his rectangular garden plot along its length on the front side. The perimeter of the plot is 24 meters, and i

Naddik [55]

We know that
2l + 2w = 24

And:
l = 3w

By substitution:
2(3w) +2w = 24
6w + 2w = 24
8w = 24
w = 3 metres
l = 3 x 3 = 9 metres

Total cost = 24 x $25 = $600

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3 years ago

Read 2 more answers

Using the given information write the equation of parabolas:

lana [24]

The general form of the equation we need to find is (x - h)^2 = 4p(y- k).

The center is the distance between the directrix and focus.

So, center (h, k) = (3, 3/2) .

P = distance from center to the focus and it just so happens to be 1.5.

We now plug everything into the formula given above.

(x - 3)^2 = 4(1.5)(y - 3/2)

(x - 3)^2 = 6(y - 3/2)

Done!

8 0

2 years ago

For number 9 and 10 what do I put for the distance between the points?

olga55 [171]

Answer:

I have wrote just answers

4 0

2 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

2 years ago