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2 years ago

State the rule for image points after a reflection over the x-axis

Mathematics

1 answer:

meriva2 years ago

5 0

Answer:B

Step-by-step explanation: hope it helps..

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HEEEEEEEELLLLPPPP PLLLLAZZZZZZZZZ

OverLord2011 [107]

Answer:

The first picture is B and the second one is (0,-3)

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2 years ago

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I'm having trouble filling this chart out and need some help please.

777dan777 [17]

Answer:

Figure 1: complementary

figure 2: adjacent

figure 3: supplementary

figure5: supplementary

7 0

1 year ago

And could someone please help me with this one please

Goshia [24]

Answer:

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

here (h, k) = (5, 2), thus

y = a(x - 5)² + 2

To find a substitute (4, - 3) into the equation

- 3 = a + 2 ⇒ a = - 3 - 2 = - 5

y = - 5(x - 5)² + 2 ← in vertex form

Expand (x - 5)² and distribute by - 5

y = - 5(x² - 10x + 25) + 2

= - 5x² + 50x - 125 + 2

= - 5x² + 50x - 123 ← in standard form

8 0

2 years ago

Choose as many answers as apply. The government makes tax forms available at what three locations? libraries, grocery stores, po

Eva8 [605]

Libraries post office over the internet

3 0

2 years ago

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A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

3 years ago