2587 to the nearset thousand

How do I solve this inequality

Makovka662 [10]

3x+5x/4≤34
leat common multiple=4
12x+5x≤136
17x≤136
x≤136/17
x≤8

Answer: x≤8 or (-∞,8]

7 0

3 years ago

PLEASE HELP RIGHT AWAY

Angelina_Jolie [31]

Answer:

12 flips

Step-by-step explanation:

4 0

3 years ago

The measure of angle A in ABC is twice the measure of angle B and the measure of angle C is 20° less than the measure

diamong [38]

Answer:

m∡A = 100°

m∡B = 50°

m∡C = 30°

Step-by-step explanation:

let 'x' = m∡B; 50°

let '2x' = m∡A; 100°

let 'x-20' = m∡C; 30°

x + 2x + x - 20 = 180

4x - 20 = 180

4x = 200

x = 50

7 0

3 years ago

I dont know how to do this...<br>I need #48

agasfer [191]

Just switch the x and the 3

8 0

3 years ago

Read 2 more answers

g A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is cha

leva [86]

Answer:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

$z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502$

$p_v =2*P(Z>2.502)= 0.0123$

Comparing the p value with the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.

Step-by-step explanation:

Data given and notation

$X_{1}=410$ represent the number of people indicating that their financial security was more than fair for the recent year

$X_{2}=245$ represent the number of people indicating that their financial security was more than fair for the year before

$n_{1}=1000$ sample 1 selected

$n_{2}=700$ sample 2 selected

$p_{1}=\frac{410}{1000}=0.410$ represent the proportion estimated of indicating that their financial security was more than fair this year

$p_{2}=\frac{245}{700}=0.35$ represent the proportion estimated of indicating that their financial security was more than fair the year before

$\hat p$ represent the pooled estimate of p

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

$\alpha$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{410+245}{1000+700}=0.385$

z-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.

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502$

Statistical decision

Since is a two sided test the p value would be:

$p_v =2*P(Z>2.502)= 0.0123$

Comparing the p value with the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.

7 0

3 years ago