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harina [27]
2 years ago
9

Can you help me please???

Mathematics
1 answer:
kipiarov [429]2 years ago
7 0

Answer:

x=80

Step-by-step explanation:

x/2=0.5x

x/2-15=25

x/2=25+15

x/2=40

x/2*2=x

40*2=80

x=80

hope this hepls you :)

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Answer:

Step-by-step explanation:

5x-4=20

5x=24: x=24/5

y=3x-12

y=3(24/5)-12

y = (72-60)/5

y=12/5

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The perimeter of a rectangle is 800 yards. What are the dimensions of the rectangle if the length is 60 yards more than the widt
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The length would be 230, and the width would be 170, and since there are 4 sides there would be 2 of each. So 230x2=460 and 170x2=340. 460+340= 800
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Write three words that represent a positive number<br> ASAP PLEASE ‼️‼️‼️‼️‼️
pantera1 [17]

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Greater than zero

Explantion:

Hope it helps!!

4 0
2 years ago
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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