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2 years ago

What is true about the points on the graph of y=2x?

Mathematics

1 answer:

IgorC [24]2 years ago

8 0

Answer:

we can't see the graph. How are we supposed to know what the graph looks like

Step-by-step explanation:

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PLEASE HELP ASAP

Vilka [71]

Answer:

A, 152 cm.

Step-by-step explanation:

A rectangular pyramid has a volume equation of

V = (lwh)/3.

Therefore, when we plug the values in:

[(7)(5)(13)]/3 = V

V = 455/3

V = 151.666667

V is approximately equal to 152 cm^3.

I hope this helped! :)

5 0

1 year ago

Find the probability of exactly three successes in six trials of a binomial experiment in which the the probability of is 50%.

Lyrx [107]

Start by doing the binomial expansion of (x+y)^6 where x represents success. This is

(x^6y^0) + 6(x^5y^1) +15(x^4y^2) +20(x^3y^3) +15(x^2y^4) +6(x^1y^5) +(x^0y^6)

We are interested in the x^3y^3 term which represents exactly 3 sucesses. Since the probalbility of sucess and failure are both .5 we should be able to figure this out just using the coefficients of the terms which is

20/64 = .3125 which is 31.25%

4 0

3 years ago

Assume that y varies directly with x. If y=2/5 when x=1/3, find y when x=1/4. Enter in any fractions using a / (e.g. "1/2" is th

Oxana [17]

x=3/15 thus y=6/15 If x=1/4 then y =2/4

7 0

2 years ago

Help please I’ve been looking online for help but couldn’t find any.

alexandr1967 [171]

5 is 33 because the question ends up being 14+3+16

4 0

3 years ago

A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of siz

marin [14]

Answer:

7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

9. p=0.9452

Step-by-step explanation:

7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

-The sample mean, $\mu_x$ is calculated as:

$\mu_x=\mu=204.9, \mu_x=sample \ mean$

-The standard deviation, $\sigma_x$ is calculated as:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968$

8. For a random variable X.

-Given a X's population mean is 75.9, standard deviation is 9.6 and a sample size of 181

-#The sample mean, $\mu_x$ is calculated as:

$\mu_x=\mu\\\\=75.9$

#The sample standard deviation is calculated as follows:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136$

9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

$\mu_x=\mu=135.7$

#Sample standard deviation:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566$

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

$P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452$

Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452

7 0

3 years ago