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4 years ago

What figure is a dilation of Figure A by a factor of 2?

Mathematics

1 answer:

stepan [7]4 years ago

7 0

Answer: Bottom Right

Step-by-step explanation:

Since the scale factor is 2 all the lengths of the original should be multiplied by that scale factor. Don’t fall for the top left because 2•4 doesn’t equal 6. The top right is just completely off. The bottom left is if the factor is .5

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A textbook store sold a combined total of 460 physics and math textbooks in a week. The number of math textbooks sold was 62 les

sergij07 [2.7K]

Number of physics textbooks sold was 261 & number of maths textbooks sold was 199.

Let, the number of physics textbooks sold = x

According to the question,

The number of math textbooks sold was 62 less than the number of physics textbooks sold.

So, Number of maths textbooks sold = x-62

The textbook store sold a combined total of 460 physics and math textbooks in a week.

So, the equation becomes,

(x-62)+x = 460

⇒ x-62+x = 460

⇒ 2x-62 = 460

⇒ 2x = 460+62

⇒ 2x = 522

⇒ x = 522/2

⇒ x = 261

So, 261 physics textbooks were sold

(261-62) = 199 maths textbooks were sold.

Learn more about equation here :

brainly.com/question/27893282

#SPJ10

5 0

2 years ago

the three expressions, sin-1, cos-1, and tan-1 are called _____ trig functions and are used to find the measure of the acute ang

geniusboy [140]

Answer:

Inverse.

Step-by-step explanation:

8 0

4 years ago

Please help :)

spin [16.1K]

The ratios 84/105 and 128/160 are proportional

Solution:

Given, two ratios are $\frac{84}{105}$ and $\frac{128}{160}$

We have to give two different reasons to support the answer.

Reason 1:

Now, we know that, numerators and denominators should be proportional to be two fractions in proportion.

$\text { Then, } \frac{84}{128}=0.65625 \text { and } \frac{105}{160}=0.65625$

So numerators and denominators are in proportion, then given two fractions are proportion.

Reason 2:

Now, values of two fractions must be equal:

$\rightarrow \frac{84}{105}=\frac{128}{160} \rightarrow 84 \times 160=105 \times 128 \rightarrow 13440=13440$

So, values of two fractions are equal, then they are in proportion.

Hence, the given two fractions are in proportion.

7 0

3 years ago

Write the equation of a direct variation that passes through the point (-11, 13)

VMariaS [17]

<h2>Answer:</h2>

A direct variation equation is of the form y = m⋅x for some constant value m.

For a direct variation equation passing through

(x,y) = (-11,13)

13 = m × - 11

→ m = - 13/11.

so, as y = m⋅x

y = - 13/11x

Hence, the direct variation equation is [C] - 13/11x.

4 0

3 years ago

Sick-leave time used by employees of a firm in a course of one month has approximately normal distribution, with a mean of 200 h

Usimov [2.4K]

Answer:

a)0.62% probability that total sick leave for next month will be less than 150 hours.

b) 225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 200, \sigma = \sqrt{400} = 20$

a.Find the probability that total sick leave for next month will be less than 150 hours.

This probability is the pvalue of Z when X = 150. So:

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

$Z = \frac{150 - 200}{20}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062.

So there is a 0.62% probability that total sick leave for next month will be less than 150 hours.

b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

This is the value of X when Z has a pvalue of 0.90. So $Z = 1.28$

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

$1.28 = \frac{X - 200}{20}$

$X - 200 = 20*1.28$

$X = 225.6$

225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

6 0

4 years ago