Which fraction is equivalent to 35/21

Triangle ABC and Triangle DEF are similar. For triangle ABC, AB = 2x , AC = 24 inches and BC = 20. For triangle DEF, EF = 5 and

Norma-Jean [14]

X=8.

Following the similarity statement, we know that AC is to DF as BC is to EF. The ratio would be:

24/(x-2) = 20/5

Cross multiply:
24*5 = 20(x-2)

Use the distributive property:
120 = 20x - 40

Add 40 to both sides:
120+40 = 20x - 40 + 40
160 = 20x

Divide both sides by 20:
160/20 = 20x/20
8 = x

8 0

3 years ago

What is this. Please help!!

Mekhanik [1.2K]

Answer: 8x+24

Step-by-step explanation:

8 0

3 years ago

Find the missing side lengths.<br> 7<br> 5<br> 60°<br> 30°

wel

60

Step-by-step explanation:

6x10=60

5 0

3 years ago

The mean consumption of water per household in a city was 1425 cubic feet per month. Due to a water shortage because of a drough

liberstina [14]

Answer:

a) $t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

b) For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

Step-by-step explanation:

Information given

$\bar X=1175$ represent the sample mean for the cubic feets of households

$\sigma=250$ represent the population standard deviation

$n=100$ sample size

$\mu_o =1425$ represent the value to verify

$\alpha=0.025$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test that the mean consumption of water per household has decreased due to the campaign by the city council, the system of hypothesis would be:

Null hypothesis: $\mu \geq 1425$

Alternative hypothesis: $\mu < 1425$

Since we don't know the deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

Part b

For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

6 0

3 years ago

A population of voters contains 40% republicans and 60% democrats. it is reported that30% of the republicans and 70% of the demo

Blizzard [7]

To get the probability of selecting a person who is a democrat and is favour to the issue, let us first calculate for the portions in each party who is favour on the issue.

p (rep & favour) = 0.4 * 0.3 = 0.12

p (dem & favour) = 0.6 * 0.7 = 0.42

Therefore the total portions who are favour is:

p (favour) = 0.12 + 0.42 = 0.54

Hence the probability is:

P (dem & favour) = 0.42 / 0.54 = 0.7778 = 0.78

Therefore there is about 78% probability that the person selected who is favour to the issue is a democrat.

5 0

3 years ago