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suter [353]
3 years ago
7

Listed below are time intervals​ (min) between eruptions of a geyser. Assume that the​ "recent" times are within the past few​ y

ears, the​ "past" times are from around 20 years​ ago, and that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Does it appear that the mean time interval has​ changed? Is the conclusion affected by whether the significance level is 0.10 or 0.01​? Recent 78 91 89 78 58 99 62 87 70 89 82 84 56 81 75 101 61 Past 90 87 93 94 65 86 84 92 88 91 91 91

Mathematics
1 answer:
Shalnov [3]3 years ago
7 0

Answer:

p value = 0.039

t = - 2.169

Step-by-step explanation:

Applying the null and alternate hypothesis

H_{0} : U1 = U2

H_{\alpha } : U1 \neq U2

using excel worksheet to calculate for ( t and p )

t = -2.169

p = 0.039

from the results obtained

The conclusion is affected by the significance level because : 0.1 < p > 0.01

so when the significance level  is = 0.1 the Null hypothesis is rejected and we can say the mean time interval will change  while

if the significance level = 0.01 the Null hypothesis is accepted and we can not say the mean time interval has changed because the p -value is greater than 0.01

attached is the excel solution

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which equation represents an exponential function that passes through the point (2, 80)? f(x) = 4(x)5 f(x) = 5(x)4 f(x) = 4(5)x
pav-90 [236]

we know that

if the exponential function passes through the given point, then the point must satisfy the equation of the exponential function

we proceed to verify each case  if the point (2,80) satisfied the exponential function

<u>case A</u>  f(x)=4(x^{5})

For x=2 calculate the value of y in the equation and then compare with the y-coordinate of the point

so

f(2)=4(2^{5})=128

128\neq 80

therefore

the exponential function  f(x)=4(x^{5}) not passes through the point (2,80)

<u>case B</u>  f(x)=5(x^{4})

For x=2 calculate the value of y in the equation and then compare with the y-coordinate of the point

so

f(2)=5(2^{4})=80

80=80

therefore

the exponential function f(x)=5(x^{4}) passes through the point (2,80)

<u>case C</u>  f(x)=4(5^{x})

For x=2 calculate the value of y in the equation and then compare with the y-coordinate of the point

so

f(2)=4(5^{2})=100

100\neq 80

therefore

the exponential function f(x)=4(5^{x}) not passes through the point (2,80)

<u>case D</u>  f(x)=5(4^{x})

For x=2 calculate the value of y in the equation and then compare with the y-coordinate of the point

so

f(2)=5(4^{2})=80

80=80

therefore

the exponential function f(x)=5(4^{x}) passes through the point (2,80)

therefore

<u>the answer is</u>

f(x)=5(x^{4})

f(x)=5(4^{x})

7 0
3 years ago
Evaluate the expression when g=7 and h= 5 8h + g
trapecia [35]

Answer: 47

Step-by-step explanation:

we simply substitute g for 7 and h for 5.

8(5) + 7

40 + 7

47

7 0
3 years ago
Read 2 more answers
I need help finding our the perimeter
barxatty [35]

Answer:

11x + 10y - 2w

Step-by-step explanation:

Hello!

To solve for the perimeter, we add up the like terms.

What are like terms?

Like terms are terms with the same variable and degree. 4x and 5y are NOT like terms because the variable is not the same. However, 4x and 3x are like terms, as adding them gives us 7x.

4x and 4x² are not like terms, because the degree of 4x is 1 (degree means the largest exponent), but the degree of 4x² is 2.

Solve for Perimeter:

Combine the like terms by adding them up.

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  • P = 8x + 3x + 3y + 7y - 4w + 2w
  • P = 11x + 10y - 2w

The perimeter is 11x + 10y - 2w

3 0
2 years ago
Find the probability that a data value picked at random from a normally distributed population will have a standard score that c
DerKrebs [107]
P(z < 3)
From the normal distribution table,
P(z < 3) = 0.9987
8 0
3 years ago
The Gophers won 2/3 of their games. The Rattlers
ZanzabumX [31]

Answer:

48

Step-by-step explanation:

The silmutenous equations

2/3x + 1/2y = 56

X=y

4 0
3 years ago
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