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Gemiola [76]
3 years ago
7

The average number of rainy days in Seattle, Washington is listed below:

Mathematics
2 answers:
MissTica3 years ago
8 0

Aprils rainy day total would be different.

January would be 20, because 8 is bigger than 5.

February will be 20 as well, because 6 is bigger than 5.

And March would be 20 as well, because 7 is bigger than 5.

The only one that is not bigger than 5 is April, it is only 4. So it would be 10.

I hope this helps. If you have any other questions, about this or any other question, just ask. I am here to help you. As needed.

8090 [49]3 years ago
7 0
Th answer would be April
5 or above you round up (5 6 7 8 9)
4 or below you round down (4 3 2 1)
18=20
16=20
17=20
14=10
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sattari [20]

Answer:

(0, 7), (1, 11), (2, 15)

Step-by-step explanation:

to find the y values, plug in the given x values into the equation y= 4x+7

when x = 0

y = 4(0) + 7

y = 0 + 7

y = 7

so the y value that corresponds to the x value 0 is 7 which is also written as (0,7)

when x = 1

y = 4(1) + 7

y = 4 + 7

y = 11

so the y value that corresponds to the x value 1 is 11; also written as (1, 11)

when x = 2

y = 4(2) + 7

y = 8 + 7

y = 15

so the x value 2 corresponds to the y value 15; also written as (2 , 15)

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3 years ago
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How does the graph of g(x) = (x − 1)3 + 5 compare to the parent function f(x) = x3?
ser-zykov [4K]

Answer:

The original function was transformed by a a horizontal shift to the right in 1 unit, and also a vertical shift upwards of 5 units.

Step-by-step explanation:

Recall the four very important rules regarding translations (shifts) of the graph of functions:

1) In order to shift the graph of a function vertically c units upwards, we must transform  f (x) by adding c to it.

2) In order to shift the graph of a function vertically c units downwards, we must transform  f (x) by subtracting c from it.

3) In order to shift the graph of a function horizontally c units to the right, we must transform the variable x by subtracting c from x.

4) In order to shift the graph of a function horizontally c units to the left, we must transform the variable x by adding c to x.

We notice that in our case, The original function f(x)=x^3 has been transformed by "subtracting 1 unit from x", and by adding 5 units to the full function. Therefore we are in the presence of a horizontal shift to the right in 1 unit (as explained in rule 3), and also a vertical shift upwards of 5 units (as explained in rule 1).

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3 years ago
A random sample of 49 measurements from one population had a sample mean of 15, with sample standard deviation 5. An independent
kotykmax [81]

Answer:

Comparing the p value with the significance level \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.  

Step-by-step explanation:

\bar X_{1}=15 represent the mean for sample 1

\bar X_{2}=18 represent the mean for sample 2

s_{1}=5 represent the sample standard deviation for 1  

s_{f}=6 represent the sample standard deviation for 2  

n_{1}=49 sample size for the group 2  

n_{2}=64 sample size for the group 2  

\alpha=0.01 Significance level provided

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 difference in the population means, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}=0  

Alternative hypothesis:\mu_{1} - \mu_{2}\neq 0  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

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

And the degrees of freedom are given by df=n_1 +n_2 -2=49+64-2=111  

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.

With the info given we can replace in formula (1) like this:  

t=\frac{(15-18)-0}{\sqrt{\frac{5^2}{49}+\frac{6^2}{64}}}}=-2.897

P value

Since is a bilateral test the p value would be:  

p_v =2*P(t_{111}  

Comparing the p value with the significance level \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.  

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