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MariettaO [177]
3 years ago
5

State the domain and range of this relation and whether or not it is a function Thanks!

Mathematics
1 answer:
tatiyna3 years ago
5 0

Answer:

i just know is not a function

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Write a slope-intercept equation for a line passing through the point (5,-5) that is parallel to the line x = -2. Then write a s
Gnesinka [82]

Answer:

1. y=-2x+5

2. y=1/2x-7.5

Step-by-step explanation:

you plug in the cordinates for the y intercept and you already have the slope.

y=mx+b

m= slope which is -2

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Can somebody please help to answer the above question and explain please
Elza [17]
The correct answer is 11.3
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What is 3^ 81^5=3^ please
Korvikt [17]

Answer:

             

Step-by-step explanation:

             

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2 years ago
Which facts could be applied to the simplified expression. Check all that apply. 5x+3y+(-x)+62
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Answer:

4x+3y+62

Step-by-step explanation:

the other variables wouldnt be able to do anything to 62 because they are different variables.

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Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0
3 years ago
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