Ask question

Ask question

All categories

love history [14]

2 years ago

11

A portion of a hiking trail slopes upward at about a 6 angle. To the nearest tenth of a foot, what is the value of x, the distan

ce the hiker traveled along the path, if he has traveled a horizontal distance of 120 feet?

1 answer:

Firdavs [7]2 years ago

4 0

The distance covered by the hiker if he traveled a horizontal distance of 120 ft at and angle of 6° will be given by:
cos θ=adjacent/hypotenuse
let the hypotenuse be h, adjacent =120 ft, θ=6°, thus plugging in the values we shall have:
cos 6=120/h
h=120/cos 6
h=120.661 ft
Answer: 120.661 ft

You might be interested in

Lenmana started studying how the number of branches on her tree grows over time. The number of branches increases by a factor of

Ad libitum [116K]

Answer:

$N(t) = 48 * 11/6^{\frac{t}{3.5} }$

Step-by-step explanation:

The formula for <u>exponential growth</u> is y = ab^x.

To write this equation, we know it has to start with 48 (which is the variable a). We need to add the rate of growth. This is 11/6 (which is variable b). But we also need to account for the "every 3.5 years" part, so divide the x as an exponent by 3.5.

N(t) = 48 * 11/6^(t/3.5)

This equation is easy to test, and it's a good idea to test it after you write it. For example, after 3.5 years we know that it should have 48*11/6 branches. Does our equation work? Yes.

7 0

1 year ago

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 449 gram setting. It is

stealth61 [152]

Answer:

We accetp H₀

Step-by-step explanation:

Information:

Normal distribution

Population mean = μ₀ = 449

Population standard deviation σ unknown

Sample size n = 23 n < 30 we use t-student test

so n = 23 degree of fredom df = n - 1 df = 23- 1 df = 22

Sample mean μ = 448

Sample standard deviation s = 20

Significance level α = 0,05

1.-Hypothesis Test

Null hypothesis H₀ μ₀ = 449

Alternative hypothesis Hₐ μ₀ ≠ 449

Problem statement ask for determine decision rule for rejecting the null hypothesis. For rejecting the null hypothesis we have to get an statistic parameter wich implies that μ is bigger or smaller than μ₀

2.-Significance level α = 0,05 ; as we have a two tail test

α/2 = 0,025

Then from t - student table for df = 22 and 0,025 (two tail-test)

t(c) = ± 2.074

3.- Compute t(s)

t(s) = ( μ - μ₀ ) / s /√n

plugging in values

t(s) = (448 - 449) / 20 /√23 ⇒ t(s) = - 1*√23 /20

t(s) = - 0.2398

4.-Compare t(c) and t(s)

t(s) < t(c) - 0.2398 < - 2.074

Therefore t(s) in inside acceptance region. We accept H₀

7 0

2 years ago

The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi

SVETLANKA909090 [29]

Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so $n = 1210$ .

The mean of the sample is 1030.

The probability of a success is $\pi = \frac{1030}{1210} = 0.8512$ .

The standard deviation of the sample is $s = \sqrt{n\pi(1-\pi)} = \sqrt{1210*0.8512*0.1488} = 12.38$

(b) Is that county result of 1030 unusually high, or low, or neither?

The first step is find the zscore when $X = 1030$ .

Then we find the pvalue of this zscore.

If this pvalue is bigger than 0.95, the county result is unusually high.

If this pvalue is smaller than 0.05, the county result is unusually low.

Otherwise, it is neither.

The national mean is 82%. So,

$\mu = 0.82(1210) = 992.2$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1030 - 992.2}{12.38}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989.This means that the county result is unusually high.

4 0

2 years ago

What is the length of EF in the right triangle below?

Lyrx [107]

Answer:

\sqrt(217)

Step-by-step explanation:

(19)^2=(12)^2

361 - 144 + 217

\sqrt(217)

8 0

2 years ago

Read 2 more answers

A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th

GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.22, n = 425$

So

$\mu = E(X) = np = 425*0.22 = 93.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54$

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{82 - 93.5}{8.54}$

$Z = -1.35$

$Z = -1.35$ has a pvalue of 0.0885.

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

8 0

2 years ago