MATH HELP PLEASE :) NO FAKE ANSWERS PLEASE :)

An example of dependent events is drawing a blue marble out of one jar and then drawing a

Karo-lina-s [1.5K]

<u>Answer:</u>

A red marble out of the same jar, without replacing the first marble.

<u>Step-by-step explanation:</u>

Dependent events are those which involve a first event which affects the outcome of the second event.

This means that the probability of the second event changes because of the occurrence of the first event.

So if a blue marble is drawn out of one jar, then drawing a red marble out of the same jar, without replacing the first marble would be an example of dependent events.

7 0

3 years ago

Cost of meal for 3 at Olive Garden Restaurant with the tax is $54.

Lynna [10]

It should be 8 dollars

3 0

3 years ago

If -1 and 2 are the two zeros of the polynomial x3-4x2-7x+10 find the third zero

rewona [7]

note that the zeroes should be x = 1 and x = - 2 not x = - 1 and x = 2

the factors are then ( x - 1 ) and (x + 2)

and x³ - 4x² - 7x + 10 = (x - 1 )(x + 2) = x² + x - 2

thus x² + x - 2 is a factor

dividing x³ - 4x² - 7x + 10 by x² + x - 2 gives x - 5

the third zero is x = 5

check : (5)³ -4(5)² - 7(5) + 10 = 125 - 100 - 35 + 10 = 0

5 0

3 years ago

I need the result of 14?2?7?3?9=10

PilotLPTM [1.2K]

Answer: yes

Step-by-step explanation:

7 0

3 years ago

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re

Zolol [24]

Answer:

a) $df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

b) $df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

c) $df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:

Null hypothesis: $\mu \leq 20$

Alternative hypothesis: $\mu > 20$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

3 0

4 years ago