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3 years ago

Which phrase best describes the translation from the graph y = 2(x - 15)2 + 3 to the graph of y = 2(x - 11)2 + 3?

Mathematics

1 answer:

ivolga24 [154]3 years ago

8 0

Answer:

4 units to the left

Step-by-step explanation:

y = a (x - h)² + k

The h is a horizontal translation of h units.

For h > 0, move right

For h < 0, move left

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Which is equal to 47 __ 5

Sergeu [11.5K]

D
•
•
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5x9= 45 remainder3 which means it 9 3/5

5 0

3 years ago

Please answer fast !!

Alborosie

Answer:

x=2

Step-by-step explanation:

2*2-x=2

4-x=2 subtract 4 by both sides

-x= -2 multiply each side by -1

x=2 answer

3 0

2 years ago

Find the area of the triangle. Round your answer to the nearest thousandths. A=1/2 bh

Over [174]

Answer:

Step-by-step explanation:

height of triangle=√(11²-8.6²)=√(11+8.6)(11-8.6)=√(19.6×2.4)

=√(4×4.9×4×0.6)=4√(4.9×0.6)=4√2.94

area of triangle=1/2 bh=1/2×2×8.6×4√2.94

=34.4×√2.94

≈58.984

5 0

2 years ago

I need help fast what is 9m−(3+7m) simplified?

iogann1982 [59]

Answer:

Simplifying 9m + -3 + -7m = 0

Reorder the terms: -3 + 9m + -7m = 0

Combine like terms: 9m + -7m = 2m -3 + 2m = 0 Solving -3 + 2m = 0

Solving for variable 'm'. Move all terms containing m to the left, all other terms to the right.

Add '3' to each side of the equation. -3 + 3 + 2m = 0 + 3 Combine like terms: -3 + 3 = 0 0 + 2m = 0 + 3 2m = 0 + 3 Combine like terms: 0 + 3 = 3 2m = 3

Divide each side by '2'. m = 1.5

Step-by-step explanation:

Hope this helps :)

4 0

2 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago