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3 years ago

14

There are 12 red marbles in a bag. The ratio of the total number of marbles in the bag to the number of red marbles in the bag i

s 5:2. How many marbles are in the bag?

1 answer:

harina [27]3 years ago

5 0

Answer:

30 marbles

Step-by-step explanation:

x = total # marbles

5/2 = x/12

2x = 60

x = 30

You might be interested in

At the market, Ms. Winn bought 3/4 lb of grapes and 5/8 lb of cherries.

wolverine [178]

Answer:

Part a) Ms. Winn bought 12 oz of grapes

Part b) Ms. Winn bought 10 oz of cherries

Part c) Ms. Winn bought 2 oz more of grapes than cherries

Part d) Mr. Phillips bought 6 more ounces of fruit than Ms. Winn

Step-by-step explanation:

Part a) How many ounces of grapes did Ms. Winn buy?

Remember that

$1\ lb=16\ oz$

To convert lb to oz multiply by 16

we know that

Ms. Winn bought 3/4 lb of grapes

Convert lb to oz

Multiply by 16

$\frac{3}{4}\ lb= \frac{3}{4}(16)=12\ oz$

therefore

Ms. Winn bought 12 oz of grapes

Part b) How many ounces of cherries did Ms. Winn buy?

Remember that

$1\ lb=16\ oz$

To convert lb to oz multiply by 16

we know that

Ms. Winn bought 5/8 lb of cherries

Convert lb to oz

Multiply by 16

$\frac{5}{8}\ lb= \frac{5}{8}(16)=10\ oz$

therefore

Ms. Winn bought 10 oz of cherries

Part c) How many more ounces of grapes than cherries did Ms. Winn buy?

Subtract the ounces of cherries from the ounces of grapes

so

$12\ oz-10\ oz=2\ oz$

therefore

Ms. Winn bought 2 oz more of grapes than cherries

Part d) If Mr. Phillips bought 1 3/4 pounds of raspberries, who bought more fruit, Ms. Winn or Mr. Phillips? How many ounces more?

Remember that

$1\ lb=16\ oz$

To convert lb to oz multiply by 16

we know that

Mr. Phillips bought 1 3/4 pounds of raspberries

Convert mixed number to an improper fraction

$1\frac{3}{4}\ lb=\frac{1*4+3}{4}=\frac{7}{4}\ lb$

Convert lb to oz

Multiply by 16

$\frac{7}{4}\ lb= \frac{7}{4}(16)=28\ oz$

Remember that

Ms. Winn bought 12 oz of grapes and 10 oz of cherries

In total Ms. Winn bought

12 oz+10 oz=22 oz -----> total ounces of fruit

so

Mr. Phillips bought 28 oz of fruit

Ms. Winn bought 22 oz of fruit

$28\ oz > 22\ oz$

Find the difference

$(28\ oz-22\ oz)=6\ oz$

therefore

Mr. Phillips bought 6 more ounces of fruit than Ms. Winn

4 0

3 years ago

D=6, Dx=-6, Dy=-24, Dz=-30 what is the solution set

Elis [28]

Answer:

(x, y, z) = (-1, -4, -5)

Step-by-step explanation:

(x, y, z) = (Dx, Dy, Dz)/D = (-6, -24, -30)/6

(x, y, z) = (-1, -4, -5)

3 0

3 years ago

A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately

Genrish500 [490]

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

$z=\frac{\bar x-\mu}{\sigma/\sqrt N}$

where

$\bar x=mean\; of\;the \;sample$

$\mu=mean\; established\; in\; H_0$

$\sigma=standard \; deviation$

N = size of the sample.

So,

$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

$\boxed {p=0.057}$

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

<u>Step 1 </u>

Compute $\bar x_{critical}$

$1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}$

$\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721$

So <em>we would make a Type II error if our sample mean is less than 185.3721</em>.

<u>Step 2</u>

Compute the probability that your sample mean is less than 185.3711

$P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z$

So, <em>the probability of making a Type II error is 0.5234 = 52.34% </em>

c)

<em>The power of a hypothesis test is 1 minus the probability of a Type II error</em>. So, the power of the test is

1 - 0.5234 = 0.4766

3 0

3 years ago

How do I find the value of X.

lora16 [44]

X = 15 , see photo for solutions

7 0

3 years ago

According to a report from a business intelligence company, smartphone owners are using an average of 22 apps per month. Assume

Ira Lisetskai [31]

Answer:

0.4332 = 43.32% probability that the sample mean is between 21 and 22.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

According to a report from a business intelligence company, smartphone owners are using an average of 22 apps per month.

This means that $\mu = 22$

Standard deviation is 4:

This means that $\sigma = 4$

Sample of 36:

This means that $n = 36, s = \frac{4}{sqrt{36}}$

What is the probability that the sample mean is between 21 and 22?

This is the p-value of Z when X = 22 subtracted by the p-value of Z when X = 21.

X = 22

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{22 - 22}{\frac{4}{sqrt{36}}}$

$Z = 0$

$Z = 0$ has a p-value of 0.5.

X = 21

$Z = \frac{X - \mu}{s}$

$Z = \frac{21 - 22}{\frac{4}{sqrt{36}}}$

$Z = -1.5$

$Z = -1.5$ has a p-value of 0.0668.

0.5 - 0.0668 = 0.4332

0.4332 = 43.32% probability that the sample mean is between 21 and 22.

4 0

2 years ago