Share 747 in the ratio 2:7

Kayleigh has 3 1/2 gallons of water. She wants to fill up 1/3 cups. How many cups can she fill and how much is left over?

erastovalidia [21]

3 1/2 gallons into cups:

56

56 cups / 1/3:

169.69697 US cups

3 0

2 years ago

Read 2 more answers

Subtract 1/6a+3 from 1/3a−5 . Please Help Will Give !Brainliest! and Medals!!!!!!

dexar [7]

A/6 - 8

Hope this helps!

6 0

2 years ago

:))))))))))))))))))))))))))))

bixtya [17]

Answer:

12 inches

Step-by-step explanation:

Set up an equation:

Variable x = height of the scale model

3/18 = x/72

Cross multiply:

3 × 72 = 18 × x

216 = 18x

Divide both sides by 18:

12

4 0

2 years ago

The distribution of weights of potato chip bags filled off a production line is unknown. However, the mean is m=13.35 OZs and th

Nikitich [7]

Answer:

a) $\mu_{\bar X} =13.35$

e.none of the above

b) $\sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02$

a.0.0200

c) $z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5$

$P(\bar X< 13.32)= P(z$

a.0.0668

d) $z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5$

$z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5$

$P(13.30$

c.0.6853

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

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".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(13.35,0.12)$

Where $\mu=13.35$ and $\sigma=0.12$

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

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

Part a

$\mu_{\bar X} =13.35$

Part b

$\sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02$

Part c

We want this probability:

$P(\bar X< 13.32)$

For this case we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got:

$z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5$

$P(\bar X< 13.32)= P(z$

Part d

We want this probability:

$P(13.30$

For this case we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got:

$z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5$

$z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5$

$P(13.30$

8 0

3 years ago

At a dance camp, students must specialize in one style of dance. Of the 62 dancers at the

kicyunya [14]

Answer:

10/31

or

0.32

or

32%

Step-by-step explanation:

Students must specialize in one style of dance.

Last summer there was 62 dancers.

20 of the 62 specialized in modern dance.

Therefore the probability of a random chosen dancer to be a modern specification is 20/62.

The answer is:

Fraction:

10

---

31

Whole number:

0.32

Percent:

32%

7 0

2 years ago