Kendall received dollars for his birthday. Joshua received dollars for his birthday. Which of the following is true?

Express 1/3 as a repeating decimal and round to the nearest hundredth.

katrin [286]

Answer:

1/3 is .333... so you can write .33.... or .33 with a line on top of the 3's

Step-by-step explanation:

.3 is in the tenths and .33 is in the hundredths place

3 0

2 years ago

What is the length of segment AE?

nexus9112 [7]

Answer:

$AE=\frac{50}{3}\ units$

Step-by-step explanation:

step 1

In the right triangle ABC

Applying the Pythagoras Theorem fin the hypotenuse AC

$AC^{2} =AB^{2}+BC^{2}$

substitute

$AC^{2} =6^{2}+8^{2}$

$AC^{2} =100$

$AC =10\ units$

step 2

we know that

If two figures are similar, then the ratio of its corresponding sides is equal

so

$\frac{AB}{CD}=\frac{AC}{CE}$

substitute and solve for CE

$\frac{6}{4}=\frac{10}{CE}\\ \\CE=4*10/6\\ \\CE=\frac{20}{3}\ units$

step 3

Find the length of segment AE

AE=AC+CE

substitute the values

$AE=10\ units+\frac{20}{3}\ units=\frac{50}{3}\ units$

6 0

3 years ago

Can someone help me please?

morpeh [17]

Heya bro!!

------------------------------------

If uh r asking about scientific notification then x must be in lowest form but decimal point is after 1.

So correct answer is 1.04 * 10^2 .

If uh wanna check multiply it.. Uh will get 104.

Hope it helps uh!

5 0

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

Find the point where the line crosses the x-axis. y = –5x + 5

Ronch [10]

Every point on the x-axis has y-coordinate 0.
Let y = 0, and solve for x.

y = -5x + 5

0 = -5x + 5

5x = 5

x = 1

The line crosses the x-axis at x = 1.

8 0

2 years ago