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kupik [55]
3 years ago
13

Joey, Chloe and Haddie are driving go-carts around a track. Joey takes 14 minutes, Chloe takes 12 minutes and Haddie takes 10 mi

nutes to drive one lap. Suppose all three left the starting line at 8:15AM and drove at a constant speed. At what time will all three meet again?
Mathematics
1 answer:
tino4ka555 [31]3 years ago
4 0

3:15PM


The LCM of 10 12 and 14 is 420

420÷60=7

7 Hours after 8:15AM is 3:15PM


The LCM (Least Common Multiple) of two numbers is the smallest number that is a multiple of both.

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icang [17]

Answer:

CD ≠ EF

Step-by-step explanation:

Using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = C(- 2, 5) and (x₂, y₂ ) = D(- 1, 1)

CD = \sqrt{(-1+2)^2+(1-5)^2}

      = \sqrt{1^2+(-4)^2}

      = \sqrt{1+16} = \sqrt{17}

Repeat using (x₁, y₁ ) = E(- 4, - 3) and (x₂, y₂ ) = F(- 1, - 1)

EF = \sqrt{(- 1+4)^2+(-1+3)^2}

     = \sqrt{3^2+2^2}

     = \sqrt{9+4} = \sqrt{13}

Since \sqrt{17} ≈ \sqrt{13} , then CD and EF are not congruent

     

6 0
3 years ago
A waitress earned $9 per hour at her job plus an additional $75 in tips on Friday. She earned more than $150 total. Which inequa
krek1111 [17]

Answer:

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Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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
A.28.0<br><br> B.19.8<br> C.14.0<br><br> D.3.7
Lady_Fox [76]
Sin(angle) = opposite leg / hypotenuse

Sin(45) = Opposite leg / 28

Opposite Leg = 28 * Sin(45)

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7 0
3 years ago
Graph the function y = |8 - 2x| + 3. Which is a point on the function? (6.2, 7.4) (7.8, 11.2) (8.6, 6.8) (9.6, 7.3)
Fiesta28 [93]
For this case we have the following function:
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 Answer:
 
(6.2, 7.4)
 
See attached image

8 0
3 years ago
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