A machine packs 180 boxes of cereal in a half-hour. It takes 55 minutes to pack 330 boxes. True or False?

Given g(x)= -x-4, solve for x when g(x) =10

dalvyx [7]

Answer:

$10 = - x - 4 \\ 10 + 4 = - \times \\ 14 = - x \\ x = - 14$

5 0

3 years ago

A new computer can complete a complex biochemistry calculation in 7 hours less than an older-model computer. Working together, t

Alchen [17]

The rate a computer works is 1/time. Working together, you add the rates.
Let new computer be x, old computer be y.
x = y - 7

$\frac{1}{x} + \frac{1}{y} = \frac{1}{13} \\ \\ \frac{1}{y-7}+\frac{1}{y} = \frac{1}{13} \\ \\ \frac{2y-7}{y(y-7)} = \frac{1}{13} \\ \\ y(y-7) = 13(2y-7) \\ \\ y^2-33y+91 = 0 \\ \\ y = 29.96 \\ \\ x = y-7 = 29.96-7 = 22.96$
Rounded to nearest tenth gives:
x = 23 hours

6 0

3 years ago

Ruth decides to use the method of proportions and similar triangles to find the height of a tree. She measures the length of the

musickatia [10]

Answer:

c. 24 ft

Step-by-step explanation:

if the 12 inch ruler casts a 6 inch shadow (shadow has half size) then a tree with 12 feet long shadow will have a height of 24 feet

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

There's four number of sides and two number of triangles what is the sum of the angle measures

jasenka [17]

The answer should be sixteen

3 0

3 years ago