Ui5<br> -4<br> Solve each equation.<br> 1) 2a - 8a = 12

Hey can you please help me posted picture of question

zhuklara [117]

For this case we have the following polynomial:
x2 + 6x + 8
We note that the polynomial can be rewritten as:
(x + 4) (x + 2)
Answer:
A common factor binomial for this case is given by:
(x + 4)
option B

7 0

3 years ago

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The city of Hampton received money for improvements of the towns park. the park Committee put a random sample of 75 presidents f

Natasha2012 [34]

Answer:about a third of the residents prefer a park improvement of more trees

Step-by-step explanation:only 27 out of 75 would make it a third

3 0

3 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

Which is the graph of the linear inequality y<3x+1

Varvara68 [4.7K]

Answer:

This is in the form y=mx+b . m is the slope and b is the y-intercept. In this case, the slope is 3 and the y-intercept is −1 .

5 0

3 years ago

John has a bag of different colored marbles. He has 6 red marbles, 3 blue marbles, 7 yellow marbles, and 5 green marbles. if Joh

zhannawk [14.2K]

Answer:

Its a 15% chance

Step-by-step explanation:

15% is the probability of the blue marble

3 0

3 years ago

Ui5 -4 Solve each equation. 1) 2a - 8a = 12 - 8a