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Sliva [168]
4 years ago
14

Pleaseee help meeee 42 + 1 x 4

Mathematics
1 answer:
katrin2010 [14]4 years ago
4 0

Answer:

46

Step-by-step explanation:

4 x 1=4 then 42+4=46

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Find the value of x. Round to the nearest tenth <br> A. 11.0<br> B. 8.1<br> C. 15.3<br> D. 24.5
mel-nik [20]

Answer:

x = 11.0

Step-by-step explanation:

b = √c2 - a2

= √132 - 6.88895043503172

= √121.54236190368

= 11.02463

5 0
4 years ago
Select all that are true for the following quadratic: y = 2 x 2 + 3 x + 5
True [87]
B.complex/imaginary roots
5 0
3 years ago
A large university offers STEM (science, technology,engineenng. and mathematics) intemshups to women in STEM majors at the unive
ValentinkaMS [17]

Answer:

a) Probability of a randomly sampled women not being qualified for the internship = 0.223

b) Probability that at least 30 percent of the women in the sample will not meet the age requirement for the internships = 0.03216

c) A woman who does not meet the age requirement is more likely to be selected with a stratified random sample than with a simple random sample.

Step-by-step explanation:

Age | Probability

17 | 0.005

18 | 0.107

19 | 0.111

20 | 0.252

21 | 0.249

22 | 0.213

23 or older | 0.063

a) Only 20+ year olds are qualified for the internship

So, probability of being qualified for the internship = P(x ≥ 20)

Probability of not being qualified for the internship = P(x < 20) = P(x=17) + P(x=18) + P(x=19) = 0.005 + 0.107 + 0.111 = 0.223

b) According to the Central limit theorem, a sampling distribution of sample size as large as 100 selected from this population distribution will approximate a normal distribution. It also has that

Mean proportion of sampling distribution of women who do not meet the internship requirements = Population proportion of women who do not meet the internship requirements = p = 0.223

The standard deviation of the is given by

σₓ = √[p(1-p)/n]

n = sample size = 100

σₓ = √[(0.223×0.777)/100] = 0.041625833 = 0.04163

So, to obtain the probability that at least 30 percent of the women in the sample will not meet the age requirement for the internships

P(x ≥ 0.30)

We first standardize 0.30

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.30 - 0.223)/0.04163 = 1.85

The required probability

P(x ≥ 0.30) = P(z ≥ 1.85)

We'll use data from the normal probability table for these probabilities

P(x ≥ 0.30) = P(z ≥ 1.85) = 1 - P(z < 1.85)

= 1 - 0.96784 = 0.03216

c) Probability of women not meeting the internship requirements = 0.223

Probability of women meeting the internship requirements = 1 - 0.223 = 0.777

Or

Probability of women meeting the internship requirements = P(x ≥ 20)

= P(x=20) + P(x=21) + P(x=21) + P(x ≥ 23) = 0.777

But as the stratified sample only contains women who do not meet the internship requirements, it is more likely that A woman who does not meet the age requirement is selected with a stratified random sample than with a simple random sample.

Hope this Helps!!!

4 0
4 years ago
Please answer these and find the slope intercept form for each one.
damaskus [11]
1. 5/y=2/3x +3
2. y= 7/1-4
3. y=2/5x8
4. y=-3/4 +1 I could be wrong but I think this is right
5 0
3 years ago
A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e
Elodia [21]

Answer:

The critical value for this hypothesis test is 6.571.

Step-by-step explanation:

In this case the professor wants to determine whether the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes.

Then the variance will be, \sigma^{2}=(5.0)^{2}=25

The hypothesis to determine whether the population variance is less than 25.0 minutes or not, is:

<em>H</em>₀: The population variance is not less than 25.0 minutes, i.e. <em>σ²</em> = 25.

<em>Hₐ</em>: The population variance is less than 25.0 minutes, i.e. <em>σ²</em> < 25.

The test statistics is:

\chi ^{2}_{cal.}=\frac{ns^{2}}{\sigma^{2}}

The decision rule is:

If the calculated value of the test statistic is less than the critical value, \chi^{2}_{n-1} then the null hypothesis will be rejected.

Compute the critical value as follows:

\chi^{2}_{(1-\alpha), (n-1)}=\chi^{2}_{(1-0.05),(15-1)}=\chi^{2}_{0.95, 14}=6.571

*Use a chi-square table.

Thus, the critical value for this hypothesis test is 6.571.

7 0
4 years ago
Read 2 more answers
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