Help please I don't understand this question

Judd is 3 years older than Jordan. Jordan is 6 years younger than Chris. Write an algebraic expression for the sum of their ages

Cloud [144]

Answer:

تيتلتبنبزلمينبووبصحثنل

6 0

2 years ago

If (x, y) is a solution to the system of equations, what is the value of y? 1/4 x + 1/8 y = 2 1/3 x + 1/2 y = 4 A) 4 B) 6 C) -6

andre [41]

<h3>The value of y is 4</h3>

Solution:

Given system of equations are:

$\frac{1}{4}x + \frac{1}{8}y =2 ---------- eqn\ 1\\\\\frac{1}{3}x + \frac{1}{2}y = 4 -------------- eqn\ 2$

We have to find value of y

From eqn 1,

$\frac{1}{4}x + \frac{1}{8}y =2 \\\\2x + y = 2 \times 8\\\\2x + y = 16 ---- eqn\ 3$

From eqn 2,

$\frac{1}{3}x + \frac{1}{2}y = 4\\\\2x + 3y = 4 \times 6\\\\2x + 3y = 24 ------ eqn\ 4$

Subtract eqn 3 from eqn 4

2x + 3y = 24

2x + y = 16

( - ) -----------------

2y = 8

y = 4

Thus value of y is 4

5 0

2 years ago

Lisa is in charge of planning a reception for 3600 people. She is trying to decide which snacks to buy. She has asked a random s

Olenka [21]

Answer:

The expected number of the people at the reception whose favorite snack will be chips is 1152

Step-by-step explanation:

Given

The data in the above table and

Reception = 3600

Required

Predict the number of the people at the reception whose favorite snack will be chips.

The first step is to determine the total number of samples;

$Total = Brownies + Cookies + Chips + Other$

$Total = 32 + 22 + 56 + 65$

$Total = 175$

The next step is to determine the fraction of people whose favorite is chips

$Fraction = \frac{Chips}{Total}$

$Fraction = \frac{56}{175}$

The product of the above fraction and the expected number of people in the reception is the solution to this question;

$Expected\ Number = \frac{56}{175} * 3600$

$Expected\ Number = \frac{56* 3600}{175}$

$Expected\ Number = \frac{201600}{175}$

$Expected\ Number = 1152$

Hence, the expected number of the people at the reception whose favorite snack will be chips is 1152

5 0

2 years ago

suppose your grandparents started a savings acct for your on first birthday with a deposit of 100. On your second birthday they

Likurg_2 [28]

1st birthday-100
2nd birthday-200
3rd birthday- 300
4th birthday- 400
5th birthday- 500
6th birthday- 600
7th birthday- 700
8th birthday- 800
9th birthday- 900
10th birthday- 1,000

4 0

3 years ago

An aptitude test is designed to measure leadership abilities of the test subjects. Suppose that the scores on the test are norma

Komok [63]

Using the normal distribution, it is found that 0.0329 = 3.29% of the population are considered to be potential leaders.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of 550, hence $\mu = 550$ .

The standard deviation is of 125, hence $\sigma = 125$ .

The proportion of the population considered to be potential leaders is 1 subtracted by the p-value of Z when X = 780, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{780 - 550}{125}$

$Z = 1.84$

$Z = 1.84$ has a p-value of 0.9671.

1 - 0.9671 = 0.0329

0.0329 = 3.29% of the population are considered to be potential leaders.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213

8 0

2 years ago