All categories

2 years ago

Abbie is 5 years older than twice her sister's age. Abbie is 25 years old. Which equation can be used to find her sister's age?

1 answer:

5 0

First of all, we can eliminate choices B and D because subtraction will not help us find the sister's age.

Since Abbie is 5 years older than twice her sister's age, we can multiply to figure out her sister's age. To form the equation, letter C would be your best choice.

Hope that helps. -UnicornFudge aka Nadia

You might be interested in

Fruit juice comes in two types of containers. The first is a rectangular prism that is 9 inches tall and has a rectangular base

My name is Ann [436]

The volume of a rectangular prism:

$V_p=Bh\\\\B=wl\\\\w=2\ in;\ l=3\ in;\ h=9\ in$

substitute

$V_p=2\cdot3\cdot9=54\ in^3$

The volume of a cylinder:

$V_c=\pi r^2h\\\\r=1.5\ in;\ h=8\ in$

substitute:

$V_c=\pi\cdot(1.5)^2\cdot8=\pi\cdot2.25\cdot8=18\pi\ in^3\approx18\cdot3.14=56.52\ in^3$

Answer: C. The cylinder has the greater volume.

3 0

2 years ago

A man bought a mobile phone for $800 and sold it for $1000. What was his profit as a percentage of the cost price

bija089 [108]

Answer:

1000/800 = 125% profit

5 0

2 years ago

Read 2 more answers

A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well

kogti [31]

Answer:

Step-by-step explanation:

Hello!

To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:

Group 1: Ate healthy breakfast

X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.

n₁= 26

Group 2: Did not eat healthy breakfast

X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.

n₂= 20

After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:

p'₁= 0.50

p'₂= 0.40

The parameters of study are the population proportions, if the claim is true then p₁ > p₂

And you can determine the hypotheses as

H₀: p₁ ≤ p₂

H₁: p₁ > p₂

α: 0.05

$Z= \frac{(p'_1-p'_2)-(p_1-p_2)}{\sqrt{p'(1-p')[\frac{1}{n_1} +\frac{1}{n_2}] } } }$ ≈N(0;1)

pooled sample proportion: $p'= \frac{x_1+x_2}{n_1+n_2} =\frac{13+8}{46} = 0.46$

$Z_{H_0}= \frac{(0.5-0.4)-0}{\sqrt{0.46(1-0.46)[\frac{1}{26} +\frac{1}{20}] } } }= 0.67$

p-value: 0.2514

The decision rule is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.

At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.

So having a healthy breakfast doesn't seem to improve the grades of students.

I hope this helps!

5 0

2 years ago

A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr

erastovalidia [21]

Answer:

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

$p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034$

The estimated standard error is:

$\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316$

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

$LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096$

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

8 0

2 years ago

A family dines in a popular franchise restaurant. They plan to use a coupon that will give them a discount of 10% off of their t

Alecsey [184]

See you add 990 then diced by 54 and then you’ll get your answer

8 0

2 years ago