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4 years ago

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In the table below, x represents miles traveled and y represents the cost to travel by train. Miles, x 2 5 Cost, y 8.50 15.25 22

.00 31.00 8 12 What is the y-intercept of this function? 0 2.25 4.00 6.50 O 17.13

2 answers:

beks73 [17]4 years ago

5 0

Answer:

4.00

I just did the question

Orlov [11]4 years ago

3 0

Answer:

4.00

Step-by-step explanation:

I just did the question

You might be interested in

Write a word problem which can be solved using the expression 2x+50

xxMikexx [17]

Answer:

friend this is not a formula

Step-by-step explanation:

ax ± b = c. All problems like the following lead eventually to an equation in that simple form.

Jane spent $42 for shoes. This was $14 less than twice what she spent for a blouse. How much was the blouse?

Solution. Every word problem has an unknown number. In this problem, it is the price of the blouse. Always let x represent the unknown number. That is, let x answer the question.

Let x, then, be how much she spent for the blouse. The problem states that "This" -- that is, $42 -- was $14 less than two times x.

Here is the equation:

2x − 14 = 42.

2x = 42 + 14

= 56.

x = 56

2

= 28.

The blouse cost $28.

7 0

3 years ago

A nutritionist has developed a diet that she claims will help people lose weight. Twelve people were randomly selected to try th

deff fn [24]

The diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

<h3>When do we use two-sample t-test?</h3>

The two-sample t-test is used to determine if two population means are equal.

A nutritionist has developed a diet that she claims will help people lose weight. In this,

Twelve people were randomly selected to try the diet.
Their weights were recorded prior to beginning the diet and again after 6 months.

Here are the original weights, in pounds, with the weight after 6 months in parentheses.

Before 192 212 171 215 180 207 165 168 190 184 200 196
After 183 196 174 211 160 191 162 175 190 179 189 195

The mean of the weights before 6 moths is,

$\overline X_1=\dfrac{192+ 212 +171 +215 +180 +207 +165 +168 +190 +184 +200 +196 }{12}\\\overline X_1=190$

The mean of the weights after 6 months is,

$\overline X_2=\dfrac{ 183 +196 +174 +211 +160 +191 +162 +175 +190 +179 +189 +195 }{12}\\\overline X_1=183.75$

Standard deviation of both the data is 16.9 and 14.7.

1. Null and Alternative Hypotheses.

The following null and alternative hypotheses need to be tested:

$\begin{array}{ccl} H_0: \mu_1 & = & \mu_2 \\\\ \\\\ H_a: \mu_1 & > & \mu_2 \end{array}$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05 and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

$df_{Total} = df_1 + df_2 = 11 + 11 = 22$

Hence, it is found that the critical value for this right-tailed test is

$t_c=1.717$ , for α=0.05 and df=22

The rejection region for this right-tailed test is,

$R = \{t: t > 1.717\}$

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$

$\displaystyle \frac{ 190 - 183.75}{\sqrt{ \frac{(12-1)16.9^2 + (12-1)14.7^2}{ 12+12-2}(\frac{1}{ 12}+\frac{1}{ 12}) } } = 0.967$

(4) Decision about the null hypothesis

Since it is observed that t=0.967≤tc=1.717 it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1721 and since p=0.1721≥0.05p = 0.1721 it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is greater than μ2, at the α=0.05 significance level.

Confidence Interval

The 95% confidence interval is −7.16<μ<19.66

Thus, the diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

Learn more about the two-sample t-test here;

brainly.com/question/27198724

#SPJ1

8 0

2 years ago

If X is a normal random variable with parameters µ = 10 and σ 2 = 36, compute

alex41 [277]

Answer:

Step-by-step explanation:

Given that X is a normal random variable with parameters µ = 10 and σ 2 = 36,

X is N(10, 6)

Or z = $\frac{x-10}{6}$

is N(0,1)

a) P(X > 5),

= $P(Z>-0.8333)\\=0.7977$

(b) P(4 < X < 16),

= $P(|z|$

(c) P(X < 8),

= $P(Z$

(d) P(X < 20),

= $P(Z$

(e) P(X > 16).

=P(Z>-0.6667)

= 0.2524

6 0

3 years ago

PLEASE HELP! I'm struggling on questions like this one. If you can please help on other questions but that's up to you! I will g

Setler79 [48]

Answer:

y=3/2x+3

Step-by-step explanation:

the line hits the y-axis at 3 so b=3 and the change in y is up 3 and the change in x is to the right 2 so the change in y over the change in x is 3/2

7 0

3 years ago

Calcium is essential to tree growth because it promotes the formation of wood and maintains cell walls. In 2010, the concentrati

sergij07 [2.7K]

Answer:

The p-value obtained from the hypothesis test is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis & say that there isn't enough evidence from the sample to conclude that the the calcium concentrations have changed since 2010.

Step-by-step explanation:

To perform this test, we need to first obtain the sample mean and sample standard deviation.

The data is 0.065 0.087 0.070 0.262 0.126 0.183 0.120 0.234 0.313 0.108

Mean = (sum of variables)/(sample size)

= (0.065+0.087+0.070+0.262+0.126+0.183+0.120+0.234+0.313+0.108)/10

= 0.1568 milligram per liter

Standard deviation = σ = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = 0.00842724+0.00487204+0.00753424+0.01106704+0.00094864+0.00068644+0.00135424+0.00595984+0.02439844+0.00238144

= 0.0676296

σ = √[Σ(x - xbar)²/N] = √(0.0676296/10) = 0.08224 milligrams per liter

To do an hypothesis test, we need to first define the null and alternative hypothesis

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, the null hypothesis would be that there is no significant evidence to suggest that the the calcium concentrations have changed since 2010. That is, the calcium concentrations haven't changed since 2010.

The alternative hypothesis is that there is significant evidence to suggest that the the calcium concentrations have changed since 2010. That is, the calcium concentrations haven't changed since 2010.

Mathematically,

The null hypothesis is represented as

H₀: μ = 0.11 milligrams per liter

The alternative hypothesis is given as

Hₐ: μ ≠ 0.11 milligrams per liter

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 0.1568 milligram per liter

μ₀ = the mean calcium concentrations from 2010 = 0.11 milligrams per liter

σₓ = standard error = (σ/√n)

where n = Sample size = 10

σ = Sample standard deviation = 0.08224 milligrams per liter

σₓ = (σ/√n) = (0.08224/√10) = 0.026

t = (0.1568 - 0.11) ÷ 0.026

t = 1.80

checking the tables for the p-value of this t-statistic

Degree of freedom = df = 10 - 1 = 10 - 1 = 9

Significance level = 0.05

The hypothesis test uses a two-tailed condition because we're testing in two directions. (Greater than and less than)

p-value (for t = 1.80, at 0.05 significance level, df = 9, with a two tailed condition) = 0.105391

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.105391

0.105391 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there isn't enough evidence to conclude that the the calcium concentrations have changed since 2010.

Hope this Helps!!!

8 0

3 years ago