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madreJ [45]
3 years ago
13

Shane set his watch 18 seconds behind, and it falls behind another 2 seconds every day. How many days has it been since Shane la

st set his watch if the watch is 44 seconds behind?
Mathematics
1 answer:
suter [353]3 years ago
7 0

Answer:

14 days

Step-by-step explanation:

18 seconds on day 1

44-18=26

26/2=13 days

Day 1 +13 Additional days=14 days

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The graph shows the functions f(x), p(x), and g(x):
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Part can’t get what ya want sooo idk what to tell u sorry not sorry
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3 years ago
Suppose you are a climatologist. You conduct a hypothesis test to determine whether the global mean temperature in the current y
just olya [345]

Answer:

The term "El Nino" refers to the warming of the central and eastern tropical Pacific waters that occurs every 3 to 7 years and typically lasts from 9 to 12 months. The 1997-1998 El Nino, the strongest ever recorded, affected climate patterns worldwide. Its effect, combined with an increasing trend in annual global temperatures, made 1998 the warmest year in the 20th century. Suppose you are a climatologist. You conduct a hypothesis test to determine whether the global mean temperature in the current year is different from the global mean temperature in 1998. Assume that the global mean temperature in 1998 is 14.3 degrees Celsius. You obtain a preliminary sample of temperatures from recording stations worldwide, which yields a sample mean of x bar = 15.1 degrees Celsius. Let mu denote the global mean temperature in the current year. Formulate your null and alternative hypotheses by selecting the appropriate values in the blue drop-down menus that follow.

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6 0
2 years ago
The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call in and get answers to questions as they
sukhopar [10]

Answer:

From the result of the one-tailed z-test, we can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Step-by-step explanation:

Sample size = 50

Mean waiting time = 13 minutes

Standard deviation = 11 minutes

We need to perform one hypothesis test that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.

The null hypothesis would be that there is no significant difference.

H₀: μ₀ = 15 minutes

The alternative hypothesis would be that the mean waiting time is indeed significantly less than the 15-minute claim by the taxpayer advocate.

Hₐ: μ₀ < 15 minutes.

This is evidently a one tail hypothesis test (we're investigating only in one direction; less than the claim). Hence, we can use the z-test

z = (x - μ)/σₓ

σₓ = (σ/√n) = (11/√50) = 1.556

z = (13 - 15)/1.556 = - 1.29

Using the z-table at significance level of 0.05.

p-value = 1 - 0.9115 (from the z-tables)

p-value = 0.0885

Since the p-value is more than the significance level (0.0885 > 0.05), we do not reject the null hypothesis.

Hence, we accept the null hypothesis and can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Hope this Helps!!!

4 0
3 years ago
Maggie's brother is 8 years younger than twice her age. The sum of their ages is 25.
g100num [7]

Answer:

the sum is 25 not 24?

Step-by-step explanation:

6 0
3 years ago
Evaluate the piecewise function at the given values of the independent variable
snow_lady [41]
A) For f(-1) we use the part of the piecewise function for when x < 0 because -1 < 0. Plug in -1 for x:

2(-1) + 3 = 1

b) For f(0), we use the part for when x \geq 0.

4(0) + 5 = 5

c) For f(2) we use the part for when x \geq 0.

4(2) + 5 = 13
3 0
3 years ago
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