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3 years ago

Helppppppop meeeeeee

Mathematics

2 answers:

Xelga [282]3 years ago

5 0

Reflection in the y axis I think

Alex73 [517]3 years ago

3 0

Answer:

option b

is the answer

please check it

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Unit Rate, Help me pleaseeeeee

Burka [1]

Do you need a unit rate? Or do you have an attatchment for me to solve the unit rate? If you need an example of a unit rate one could be:

27 gallons.
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27 divided by 3 is 9

9 gallons
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3 years ago

Two less than 2 times a number is the same as the number plus 64. What is the number?

Alex Ar [27]

Set it up as an equation. "A number" is x.

2x-2=x+64

Move all of the x terms to one side by subtracting x from both sides, then add 2 to both sides to isolate x.

X=66

The number is 66

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3 years ago

Can u help me plss i’ll give u brainliest

svet-max [94.6K]

Answer:

The answer is

Step-by-step explanation:

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3 years ago

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julia-pushkina [17]

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3 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago