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3 years ago

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Meg is a veterinarian. She found that 7,or 25% of the dogs she saw this week were boxers. Steve is also a veterinarian. He found

that 50% of the 10 dogs he saw this week were boxers. Does each person need to find the part, the whole, or the percent?
HELP NOW!!!!!!!!!

1 answer:

Alex787 [66]3 years ago

5 0

Answer:

whole

Step-by-step explanation:

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Mr Teo wants to buy an oven. Which shop sells the oven at a lower price? Show your working clearly.

Oliga [24]

Answer:

Shop B

Step-by-step explanation:

Hi there!

To solve this question, we can find the new prices of each oven and identify which one is cheaper.

<u>Shop A</u>

Usual price: $190

Discount: 15%

First, we must subtract the discount percent from 100:

100 - 15 = 85

Therefore, the new price of the product will be 85% of the original price. Find 85% of $190:

190 × 0.85

Therefore, the new price is $161.50.

<u>Shop B</u>

Usual price: $200

Discount: 20%

Again, subtract 20 from 100:

100 - 20 = 80

This means that the new price of the oven is 80% of the original price:

200 × 0.8 = 160

Therefore, the new price is $160.

Because a $160 oven is cheaper than a $161.50 oven, Shop B sells the oven at a lower price.

I hope this helps!

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3 years ago

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The length of AB¯¯¯¯¯ is 10 inches. A dilation with a scale factor of 25 is applied. What is the length of the image of AB

DaniilM [7]

The answer is 4.

I hope this helps. :)

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One<br> Find the product:<br> (n - 5)(n-1)

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The answer is n^2-6n+5

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A new phone system was installed last year to help reduce the expense of personal calls that were being made by employees. Befor

Leya [2.2K]

Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 700, \sigma = 50$ .

The probability of a month having a PCE between $575 and $790 is the <u>p-value of Z when X = 790 subtracted by the p-value of Z when X = 575</u>, hence:

X = 790:

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

$Z = \frac{790 - 700}{50}$

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

X = 575:

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

$Z = \frac{575 - 700}{50}$

Z = -2.5

Z = -2.5 has a p-value of 0.0062.

0.9641 - 0.0062 = 0.9579.

0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

More can be learned about the normal distribution at brainly.com/question/4079902

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2 years ago

Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F

Svet_ta [14]

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

$y = 2x + 12$

The two graphs intersect when:

$x^{2} + 2x + 3 = 2x + 12$

$x^{2} = 0$

$x_{1} = 3\\x_{2} = -3$

To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).

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3 years ago