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3 years ago

Please I need help on this problem this assignment due tomorrow!!!!!!! be brainliest if help me

Mathematics

1 answer:

madreJ [45]3 years ago

6 0

Answer:

don't click that guy's link it's a virus

I'm pretty sure the answer if 50% chance

Step-by-step explanation:

I just combined the probabilities because it's for both lights

hope it helps <3

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What is the solution to the inequality 5.25 – b ≥ 6.5?

natita [175]

$- b \geqslant 6.5 - 5.25 \\$

$b \leqslant 1.25$

7 0

3 years ago

Read 2 more answers

i need help. Semir, Sarah, and SungSo decided to raise money for a local homeless shelter by working in a local deli. The deli a

IgorLugansk [536]

According to the question,
Let,
"n" represent the number of miles semir walked.
"y" represent the number of miles sarah walked.
Now, according to the question,

y = 2n - 5 ........................this is your equation
Also,
the question states, each of them collect $18 in pledges for every miles walked.

Given,
Sarah collected $450
Now,

Using unitary method,

Sarah collects $18 for 1 mile
Sarah collects $1 for (1 / $18) mile
Sarah collects $450 for (1 / 18) * 450 mile
= 25 miles
So, Sarah walks 25 miles.
Now,
Taking equation,
y = 2n - 5
Since, y is the no. of miles sarah walked, we can write 25 in place of "y" So,
(25) = 2n - 5
25 + 5 = 2n
30 = 2n
30 / 2 = n
15 = n

Since, "n" is the no. of miles that semir walked, Semir walked 15 miles.

7 0

3 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: