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2 years ago

Blue's Berry Farm charges Percy a total of

Mathematics

1 answer:

kaheart [24]2 years ago

8 0

Answer:

The price for each kilogram of strawberries is $7.50

Step-by-step explanation:

<u><em>The question is</em></u>

Blues berry farm charges Percy a total of $24.75 for entrance and 2.5 kilograms of strawberries. The entrance fee is $6 and the price for each kilogram of strawberries is constant

Determine the price for each kilogram of strawberries

Let

x ----> the price for each kilogram of strawberries

we know that

The entrance fee plus the price of each kilogram of strawberries multiplied by the number of kilogram of strawberries must be equal to $24.75

The linear equation that represent this situation is

$2.5x+6=24.75$

solve for x

subtract 6 both sides

$2.5x=24.75-6\\2.5x=18.75$

divide by 2.5 both sides

$x=\$7.50$

therefore

The price for each kilogram of strawberries is $7.50

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When the net is folded into the rectangular prism shown beside it, which letters will be on the front and back of the rectangula

earnstyle [38]

<h2>Answer:</h2>

<u>The correct option is </u><u>The letter on the front will be N. The letter on the back will be L. </u>

<h2>Step-by-step explanation:</h2>

When we fold the given net, we will get Q,P,M and N on sides. Side M will come to the top, side Q on the right side, side P on the left and side O on the bottom. The side which comes to the front will be N of the observer and similarly the side L will come to the back of the rectangular prism.

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2 years ago

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Ghella [55]

B) ASA Congruence

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2 years ago

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The national average for mathematics SATs in 2011 was 514 and the standard deviation was approximately 40. a) Within what bounda

blondinia [14]

Answer:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

$\mu = 514 , \sigma =40$

For this case we can use the Chebysev theorem who states that "At least $1 -\frac{1}{k^2}$ of the values lies between $\mu -k\sigma$ and $\mu +k\sigma$ "

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

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2 years ago

Find the center of the circle: x2 + 2x - 3 + y2 = 5.

rosijanka [135]

Answer:

$\boxed{(-1, 0)}$

Step-by-step explanation:

x² + 2x - 3 + y² = 5

Strategy:

Convert the equation to the centre-radius form:

(x - h)² + (y - k)² = r²

The centre of the circle is at (h, k) and the radius is r .

Solution:

Move the number to the right-hand side.

x² + 2x + y² = 8

Complete the square for x

(Take half the coefficient of x, square it, and add to each side of the equation)

(x² + 2x + 1) + y² = 9

Complete the square for y

The coefficient of y is zero.

(x² + 2x + 1) + y² = 9

Express the result as the sum of squares

(x + 1)² + y² = 3²

h = -1; k = 0; r = 3

The centre of the circle is at $\boxed{(-1,0)}$

The graph of the circle below has its centre at (-1,0) and radius 3.

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2 years ago

In a survey of men aged 20-29 in a certain country, the mean height is 73.4 inches with a standard deviation of 2.7 inches. Find

satela [25.4K]

Answer:

The minimum height in the top 15% of heights is 76.2 inches.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 73.4, \sigma = 2.7$