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Maurinko [17]
3 years ago
9

-2(a-8)+6a=36 what's the value of a A=

Mathematics
1 answer:
ivann1987 [24]3 years ago
4 0
Distribute -1 into the brackets.
-2(a - 8) + 6a = 36
-2a + 16 + 6a = 36
Take 16 to the other side
-2a + 6a + 16 = 36
               - 16   -16
-2a + 6a = 36 - 16
Combine like terms
4a = 20
Divide by 4 on either sides to isolate a
\frac{4a}{4} =  \frac{20}{4}
4 and 4 cancels out.
a = 5
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Step-by-step explanation:

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How many terms are in the expression shown? <br><br> 2n + 5 – 3p + 4q<br><br> 1<br> 2<br> 3<br> 4
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Factor completely 3bx^2 - 9x^3 - b + 3x​
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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
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Answer:

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

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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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