Please help ASAP!! Please

Look at image below :D

Artyom0805 [142]

Answers:

The answer is in the picture!

Please give brainliest. :)

7 0

2 years ago

At a fast food restaurant, three chicken sandwiches and two small orders of fries cost $8.87. Five chicken sandwiches and four s

quester [9]

Answer:

The cost of each chicken sandwich is $2.19 and the cost of each small orders of fries is $1.15.

Step-by-step explanation:

Given:

The cost of three chicken sandwiches and two small orders of fries is $8.87. And the cost of five chicken sandwiches and four small orders of fries is $15.55.

Now, to find the cost of each chicken sandwiches and each small orders of fries.

Let the cost of each chicken sandwich be $x.$

And let the cost of each small orders of fries be $y.$

So, the total cost of three chicken sandwiches and two small orders of fries:

$3x+2y=8.87$ ......(1)

Now, the total cost of five chicken sandwiches and four small orders of fries:

$5x+4y=15.55$ .......(2)

Now, to solve the equations by using elimination method:

$3x+2y=8.87$ ....(1)

$5x+4y=15.55$ .....(2)

So, multiplying equation (1) by 2 we get:

$6x+4y=17.74$

And, then subtracting equation (2) from the resulting equation (1):

$(6x+4y)-(5x+4y)=17.74-15.55\\\\6x+4y-5x-4y=2.19\\\\6x-5x+4y-4y=2.19\\\\x=2.19.$

The cost of each chicken sandwich = $2.19.

Now, substituting the value of $x$ in equation (1):

$3x+2y=8.87\\\\3(2.19)+2y=8.87\\\\6.57+2y=8.87$

Subtracting both sides by 6.57 we get:

$2y=2.3$

Dividing both sides by 2.3 we get:

$y=1.15.$

The cost of each small orders of fries = $1.15.

Therefore, the cost of each chicken sandwich is $2.19 and the cost of each small orders of fries is $1.15.

4 0

4 years ago

An article reported that in a large study carried out in the state of New York, approximately 60% of the study subjects lived wi

DerKrebs [107]

Answer:

95.86%.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 0.6, \sigma = \sqrt{\frac{p(1-p)}{n}} = 0.0245$

$P(0.55 \leq x \leq 0.65$

This probability is the pvalue of Z when $X = 0.65$ subtracted by the pvalue of Z when $X = 0.55$ . So

X = 0.65

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

$Z = \frac{0.65 - 0.6}{0.0245}$

$Z = 2.04$

$Z = 2.04$ has a pvalue of 0.9793

X = 0.55

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

$Z = \frac{0.55 - 0.6}{0.0245}$

$Z = -2.04$

$Z = -2.04$ has a pvalue of 0.0207

So this probability is 0.9793 - 0.0207 = 0.9586 = 95.86%.

5 0

3 years ago

Tom had 114 baseball cards. He kept 10 and shared the rest evenly among his 8 friends. How many baseball cards did each friend g

SVETLANKA909090 [29]

104/8=13
each friend gets 13 cards

3 0

3 years ago

Read 2 more answers

The sum of the ages of Berma, her mother Rinna and her father Erwin is 80. Two years from now, Rinna’s age will be 13 less than

jekas [21]

Answer: Berma is 5 years old

Rinna is 38 years old

Erwin is 37 years old

Step-by-step explanation:

Let x represent Berma's age

Let y represent Rinna's age

Let z represent Erwin's age

Since the sum of their ages is 80,

x + y + z = 80 - - - - - - -1

Two years from now, Rinna’s age will be 13 less than the sum of Erwin’s age and twice Berma’s age. This means that

y +2 = [ (z+2) + 2(x+2) ] - 13

y +2 = z + 2 + 2x + 4 - 13

2x - y + z = 13 + 2 - 4 -2

2x - y + z = 9 - - - - - - -2

Three years ago, 15 times Berma’s age was 5 less than the age of Rinna. It means that

15(x - 3) = (y - 3) - 5

15x - 45 = y - 3 - 5

15x - y = - 8 + 45

15x - y = 37 - - - - - - - -3

From equation 3, y = 15x - 37

Substituting y = 15x - 37 into equation 1 and equation 2, it becomes

x + 15x - 37 + z = 80

16x + z = 80 + 37 = 117 - - - - - - 4

2x - 15x + 37 + z = 9

-13x + 2 = -28 - - - - - - - - -5

subtracting equation 5 from equation 4,

29x = 145

x = 145/29 = 5

y = 15x - 37

y = 15×5 -37

y = 38

Substituting x= 5 and y = 38 into equation 1, it becomes

5 + 38 + z = 80

z = 80 - 43

z = 37

6 0

3 years ago