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Gre4nikov [31]
3 years ago
13

Tanya kicks a ball into the air. The function that models this scenario f(t)= -16t²+96t, where h is the height of the ball in fe

et and t is the time in seconds. When will the ball hit the ground?
Mathematics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

The ball hit the ground at t=6 seconds

Step-by-step explanation:

we have

f(t)=-16t^2+96t

where

f(t) is he height of the ball in feet

t is the time in seconds

we know that

When the ball hit the ground, the height of the ball is equal to zero

so

For f(t)=0

-16t^2+96t=0

solve for t

Factor the leading coefficient -16t

-16t(t-6)=0

the solutions are

t=0, t=6

Remember that

For t=0 ----> f(t)=0 is the initial height of the ball

Fort=6 sec ----> f(t)=0

therefore

The ball hit the ground at t=6 seconds

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Klio2033 [76]

Answer:

For this case the statistic calculated is t= 1.37

p_v =0.17

So the p value is a very high value and using the significance level  given\alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.

Step-by-step explanation:

Data given and notation

\bar X_{A} represent the mean for the sample for A

\bar X_{B} represent the mean for the sample B

s_{A} represent the sample standard deviation for the sample A

s_{B}= represent the sample standard deviation for the sample B

n_{A}=20 sample size for the group poisoned

n_{B}=20 sample size for the group unpoisoned

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the two means are equal or not , the system of hypothesis would be:

Null hypothesis:\mu_{A} - \mu_{B} =0

Alternative hypothesis:\mu_{A} - \mu_{B} \neq 0

For this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

For this case the statistic calculated is t= 1.37

Statistical decision

The significance level is 0.1 \alpha=0.1, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

df=n_{A}+n_{B}-2

The p value on this case is given

p_v =0.17

So the p value is a very high value and using the significance level  given\alpha=0,1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.

5 0
3 years ago
Pls help me with 1 and 2 plzzzz
kupik [55]

Answer:

1. First choice, 1/2

2. First choice, 1/2

Step-by-step explanation:

3 odd numbers on a dice, so 3/6 = 1/2

12 numbers, 1- 12, the even numbers are 2, 4, 6, 8, 1-, and 12. 6 out of 12 = 6/12 = 1/2

7 0
3 years ago
In a survey of 269 college students, it is found that
Degger [83]

Answer: a) 83, b) 28, c) 14, d) 28.

Step-by-step explanation:

Since we have given that

n(B) = 69

n(Br)=90

n(C)=59

n(B∩Br)=28

n(B∩C)=20

n(Br∩C)=24

n(B∩Br∩C)=10

a) How many of the 269 college students do not like any of these three vegetables?

n(B∪Br∪C)=n(B)+n(Br)+n(C)-n(B∩Br)-n(B∩C)-n(Br∩C)+n(B∩Br∩C)

n(B∪Br∪C)=69+90+59-28-20-24+10=156

So, n(B∪Br∪C)'=269-n(B∪Br∪C)=269-156=83

b) How many like broccoli only?

n(only Br)=n(Br) -(n(B∩Br)+n(Br∩C)+n(B∩Br∩C))

n(only Br)=90-(28+24+10)=28

c) How many like broccoli AND cauliflower but not Brussels sprouts?

n(Br∩C-B)=n(Br∩C)-n(B∩Br∩C)

n(Br∩C-B)=24-10=14

d) How many like neither Brussels sprouts nor cauliflower?

n(B'∪C')=n(only Br)= 28

Hence, a) 83, b) 28, c) 14, d) 28.

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scoundrel [369]

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Elena is making greeting cards, which she will sell by the box at an arts fair. She paid $54 for
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Answer:

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