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2 years ago

What is the answer text

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

5 0

3/4 of milk: because the bigger the number the smaller the size

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The mathematics department of a college has 10 male professors, 12 female professors, 12 male teaching assistants, and 15 fem

lana [24]

We need the total count of persons = 10+12+12+15 = 49 persons

We need the count of the target group, female or teaching assistants

= (10+12) professors + 12 male teaching assistants

= 22+12

= 34 persons

Assuming equal probability of choosing anyone from the 30 persons, the

probability of choosing a professor or a male

= 34/49

(For probability calculations, try to keep a fraction for as long as you can, because fractions are exact. Decimal are frequently approximate, for example in this case, 34/49=0.693877551020...... = 0.694 approximately)

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3 years ago

Solve The Quadratic Equation by the Square Root Method.<br><br> (x - 2)^2-1=0

agasfer [191]

Download Cymath to get the answer

6 0

2 years ago

Please help I will brainliest!! Really urgent!!

asambeis [7]

Answer:1/9(x+0.1)

Step-by-step explanation:

1/9x+10/9

=1/9(x+0.1)

8 0

2 years ago

| 2x + 2y = 10<br> y = 5+x

melisa1 [442]

2x + 2(5+x)=10
2x+10+2x=10
4x+10=10
4x=0
X=0

6 0

2 years ago

Sample Size for Proportion As a manufacturer of golf equipment, the Spalding Corporation wants to estimate the proportion of gol

Dima020 [189]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56$

And rounded up we have that n=2663

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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 population proportion have the following distribution

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.025$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume an estimated proportion of $\hat p =0.5$ since we don't have prior info provided. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56$

And rounded up we have that n=2663

6 0

3 years ago