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3 years ago

the vertex form of the equation of a parabola is y=(x+3)^2+53. what is the standard for of the equation

Mathematics

1 answer:

german3 years ago

7 0

x squared plus 9 plus 53. x squared plus 62. square root both sides. x equals 7.9 or 8.
Hope this helps.

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26,) If y varies inversely as x, and y = 5 as x = 6, find y for the x-value of 10.

melisa1 [442]

Answer:

Step-by-step explanation:

the initial statement is

y ∝ 1 /x

to convert to an equation multiply by k the constant

of variation

y = k × 1 /x = k /x

to find k use the given condition

y = 5 when x = 6

y = k/ x ⇒ k = y x = 5 × 6 = 30

y = 30 /x

when

x = 10

then

y = 30 /10 = 3

4 0

3 years ago

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Will award brainlyist! 15 pts! Find the perimeter of the window to the nearest hundredth.

yawa3891 [41]

Answer:

7.71

Step-by-step explanation:

The diameter is 3, so the arc length is 180/360 * 3pi or 3pi/2. Now you add the 3 in the base, so it is 3pi/2 + 3 or approximately 7.71 (I used a calculator for that).

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3 years ago

Which data value has the highest frequency? 116 316 38 58

lara31 [8.8K]

1/16. that is the value that has the most x's on the graph.

8 0

3 years ago

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The Genetics & IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability of conceivi

Reika [66]

Answer: The test statistic needed to test this claim= 10.92

Step-by-step explanation:

We know that the probability of giving birth to a boy : p= 0.5

i..e The population proportion of giving birth to a boy = 0.5

As per given , we have

Null hypothesis : $H_0: p\leq0.5$

Alternative hypothesis : $H_a: p>0.5$

Since $H_a$ is right-tailed , so the hypothesis test is a right-tailed z-test.

Also, it is given that , the sample size : n= 291

Sample proportion: $\hat{p}=\dfrac{239}{291}\approx0.82$

Test statistic : $z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$ , where n is sample size , $\hat{p}$ is sample proportion and p is the population proportion.

$\Rightarrow\ z=\dfrac{0.82-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{291}}}\approx10.92$

i.e. the test statistic needed to test this claim= 10.92

Critical value ( one-tailed) for 0.01 significance level = $z_{0.01}=2.326$

Decision : Since Test statistic value (10.92)> Critical value (2.326), so we reject the null hypothesis .

[When test statistic value is greater than the critical value , then we reject the null hypothesis.]

Thus , we concluded that we have enough evidence at 0.01 significance level to support the claim that the YSORT method is effective in increasing the likelihood that a baby will be a boy.

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3 years ago

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1

KATRIN_1 [288]

Answer:

Step-by-step explanation:

Heres the complete question:

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1990–94, and 1995–99. If the population at the end of 1999 was 9,320:

How many people lived in the town at the beginning of 1985? (Round your answer to the nearest whole number.)

solution:

Let the population of the town at the beginning of 1985 be P. Then, given that in the first five-year period the population declined by 3.2%, i.e., 0.032, the population of the town at the end of 1989 would be

(1 – 0.032)P = 0.968P.

Again, given that in the second five-year period the population declined by 5.2%, i.e., 0.052, the population of the town at the end of 1994 would be

(1 – 0.052)(0.968P) = 0.948 x 0.968P = 0.917664P.

Finally, given that in the third five-year period the population declined by 4.7%, i.e., 0.047, the population of the town at the end of 1999 would be

(1 – 0.047)(0.917664P) = 0.874533792P.

We are given, 0.874533792P = 9320 or

P = 9320/0.874533792 = 10657.11.

Thus, 10657 people lived in the town at the beginning of 1985

3 0

3 years ago