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3 years ago

7

A walking trail is 3.8 miles long. If Gertrude walked the length of the 1 walking trail 13 times, how many miles did Gertrude wa

lk?

1 answer:

MAVERICK [17]3 years ago

5 0

Answer:

hehduhewduehusdu

Step-by-step explanation:

pay back

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Tan x (degrees) = 0.3<br> what is the value of x?

jenyasd209 [6]

X= shift tan 0.3
x=16.6992≈17

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3 years ago

A line passes through the point (4,-2) and has a slope of 1/2 what is the value of a if the point (-4,a) is also on the line

Slav-nsk [51]

Answer: The value of a is -6.

Step-by-step explanation:

To do this we need to find the y-intercept first.

-2= 4(1/2) + b

-2=2+ b

-2 -2

b= -4 so now the y-intercept is -4 so we will now have the equation y= 1/2x -4

so now put -4 into the equation for x and solve for y.

y= 1/2(-4) -4

y = -6

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2 years ago

Enter the number 46.1979 rounded to the nearest hundredth (two decimal places)

muminat

Answer:

It should be 46.20

Step-by-step explanation:

You're rounding to the nearest hundredth. So you have 46.1979 to start out with. Then you count 2 places so you end at the 1st 9. Well, behind the 9 is a 7 so you round up and get 20

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1 year ago

Please help!! ill give BRAINLIEST!!!

Alex787 [66]

1. Similar: all angles are equivalent.

2. Similar: 2 angles and 1 side are equivalent.

3. Not enough information.

4. Not enough information.

5. Not enough information.

6. False.

7. True.

8. True.

9. True.

10. False.

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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

3 years ago