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sveta [45]
3 years ago
11

Joel has an MP3 player called the Jumble. The Jumble randomly selects a song for the user to listen to. Joel's Jumble has 222 cl

assical songs, 131313 rock songs, and 555 rap songs on it.
What is \text{P(rock song or rap song})P(rock song or rap song)start text, P, left parenthesis, r, o, c, k, space, s, o, n, g, space, o, r, space, r, a, p, space, s, o, n, g, end text, right parenthesis?
Mathematics
1 answer:
mash [69]3 years ago
7 0

ANSWER

0.9

Step-by-step explanation:

no.of favourable outcome/no.of.possible outcome  

13+5=18 favourable outcome and 20 possible outcome 18/20=0.9

You might be interested in
The area of a rectangle is 32a^3b^4 square units. The length is 4a^2b. Find the width. show your work
kaheart [24]

Answer:

The answer to your question is width = 8ab³

Step-by-step explanation:

Data

Area = 32a³b⁴ u²

length = 4a²b   u

Formula

Area of a rectangle = width x length

Solve for width

                           width = \frac{Area}{length}

Substitution

                          width = \frac{32a^{3}b^{4}}{4a^{2}b}

Simplify using rules of exponents, just remember that in a division we subtract the exponents and divide the coefficients normally.

                          width = 8 a²b³

7 0
3 years ago
Read 2 more answers
Which fraction equals a repeating decimal
In-s [12.5K]
4/9.
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3 0
4 years ago
490 as a product of its prime factors
Katena32 [7]

Answer:

2×5×7²

Step-by-step explanation:

You will start by using the factor tree.

490

^

2 245

^

5 49

^

7 7

8 0
3 years ago
A recent study focused on the number of times men and women send a WeChat message in a day. The information is summarized next.
Ratling [72]

Answer:

1)Null hypothesis:\mu_{men}=\mu_{women}

Alternative hypothesis:\mu_{men} \neq \mu_{women}

2) Two critical values are z_{\alpha/2}=-2.58 and z_{1-\alpha/2}=2.58

3) z=-2.40  

4) p_v =2*P(z

5) Comparing the p value with the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the mean number of times men and women send a Twitter message in a day.

Step-by-step explanation:

Data given and notation

\bar X_{men}=23 represent the mean for the sample men

\bar X_{women}=28 represent the mean for the sample women

\sigma_{men}=5 represent the population standard deviation for the sample men

\sigma_{women}=10 represent the population standard deviation for the sample women

n_{men}=25 sample size for the group men

n_{women}=30 sample size for the group women

t would represent the statistic (variable of interest)

\alpha=0.01 significance level provided

1)Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:\mu_{men}=\mu_{women}

Alternative hypothesis:\mu_{men} \neq \mu_{women}

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

z=\frac{\bar X_{men}-\bar X_{women}}{\sqrt{\frac{\sigma^2_{men}}{n_{men}}+\frac{\sigma^2_{women}}{n_{women}}}} (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

2)Determine the critical value(s).

Based on the significance level\alpha=0.01 and \alpha/2=0.005 we can find the critical values with the normal standard distribution, we are looking for values that accumulates 0.005 of the area on each tail on the normal distribution.

For this case the two values are z_{\alpha/2}=-2.58 and z_{1-\alpha/2}=2.58

3) Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

z=\frac{23-28}{\sqrt{\frac{5^2}{25}+\frac{10^2}{30}}}}=-2.40  

4)What is the p-value for this hypothesis test?

Since is a bilateral test the p value would be:

p_v =2*P(z

5)Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the mean number of times men and women send a Twitter message in a day.

4 0
3 years ago
Will give brainliest!!!
Eva8 [605]

Answer:

4/16

Step-by-step explanation:

brainliest pls

4 0
3 years ago
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