Age of Soccer Club Players

Dolve for theta. С 13 4 o o A B Your answer

ANTONII [103]

Answer:

θ ≈ 17.1°

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan θ = $\frac{opposite}{adjacent}$ = $\frac{BC}{AC}$ = $\frac{4}{13}$ , then

θ = $tan^{-1}$ ( $\frac{4}{13}$ ) ≈ 17.1° ( to the nearest tenth )

5 0

3 years ago

ANSWER AND EXPLAIN HOW U GOT IT=BRAINLIST FOR U:) THX

Evgesh-ka [11]

The answers r online for that

4 0

3 years ago

Sara selects two of the seven colors of the rainbow, one after another, with replacement. What is the probability that the first

Triss [41]

Well, there are actually millions of colors in the rainbow, (every color that the human eye can see), but I do understand the question you're asking:

-- When Sara is ready to make her first selection, there are seven things with different colors in the pot, and one of them is red.

The probability that she picks the red one is ( 1 / 7 ).

The question says "with replacement". That means that after she selects it and looks at it, she puts it back in the pot. So . . .

-- When she's ready to make her second selection, the same seven things with different colors are still in the pot. One of them is orange.

The probability that she picks the orange one is ( 1 / 7 ).

-- The probability of BOTH selections being successful (the color she wants) is

(1/7) x (1/7) .

That's (1/49), which is about 2.04 percent .

3 0

3 years ago

What is the polar form of 7 StartRoot 3 EndRoot minus 7 i?

ololo11 [35]

Answer:

14(cos(330 degrees) + i sin(330 degrees))

option D

Step-by-step explanation:

edge 2020

3 0

3 years ago

4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B

AleksAgata [21]

Answer:

$z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123$

$p_v =P(Z>2.123)=0.0169$

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

$\hat p=\frac{63}{200}=0.315$ estimated proportion of lawyers had used some form of advertising for their business

$p_o=0.25$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:

Null hypothesis: $p\leq 0.25$

Alternative hypothesis: $p > 0.25$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>2.123)=0.0169$

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .

8 0

4 years ago