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3 years ago

Kim works 4 hours more each day than jill does, and jack works 2 hours less each day

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

6 0

Where's the rest of the question

marusya05 [52]3 years ago

4 0

Answer:

The answer is 2(x+3)=3 * 2(x-1)-2x

Step-by-step explanation:

full question:

Kim works 3 hours more each day than Jill does, and Jack works 1 hour less each day than Jill does. Over 2 days, the number of hours Kim works is equal to the difference of 3 times the number of hours Jack works and the number of hours Jill works. How many hours does each person work each day? Let x be the number of hours Jill works in one day.

2(x+3)=3 * 2(x-1)-2x

2(x+3)=6(x-1)-2x

2x+6=4x-6

2x+6-6=4x-6-6

2x=4x-12

-2x=-12

-2x/-2=-12/-2

x=6

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Twice the larger of two consecutive integers is equal to fifteen less than three times the smaller. Find the integers.

Alexxandr [17]

Answer:

17,18

Step-by-step explanation:

let the smaller integer be "a" and the larger be "b"

Then 2b = 3a -15 ........................equ 1

but they are consecutive integers, b= a+1 ; substitute b into equ 1,thus,

2(a+1) = 3a - 15

2a + 2 = 3a-15

collect like terms,

2+15 =3a-2a

17 = a

b= a+1

b = 17+1

b=18.

The integers are 17,18

7 0

2 years ago

6 If the altitudes of a triangle meet at one of the triangle’s vertices, then the triangle is

lyudmila [28]

Answer:

1) a right triangle

Step-by-step explanation:

One of the caractheristics of a right triangle, the one in which the sum of the two sides squared is the hypotenuse squared(Pytagorean Theorem), is also the the altitudes of a triangle meet at one of the triangle’s vertices. So 1) is the answer to this question.

4 0

3 years ago

The local amusement park was interested in the average wait time at their most popular roller coaster at the peak park time (2 p

Effectus [21]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

118, 124, 108, 116, 99, 125, 138, 118, 119, 121, 65, 130, 120

Ordered data:

65, 99, 108, 116, 118, 118, 119, 120, 121, 124, 125, 130, 138

1.)

90th percentile :

0.9(n + 1)th term

n = sample size = 13

0.9(14) th term

= 12.6th term

= (12 + 13)th / 2

= (130 + 138) / 2

= 134

Third quartile (Q3) = 75th percentile

0.75(n + 1)th term

n = sample size = 13

0.75(14) th term

= 10.5th term

= (10 + 11)th / 2

= (124 + 125) / 2

= 124.5

What is the first quartile?

First quartile (Q1) :

0.25(n + 1)th term

n = sample size = 13

0.25(14) th term

= 3.5th term

= (3 + 4)th / 2

= (108 + 116) / 2

= 112

(4) What is the 10th percentile?

0.1(n + 1)th term

n = sample size = 13

0.1(14) th term

= 1.4th term

= (1 + 2)th / 2

= (65 + 99) / 2

= 82

(5) 65tb percentile :

0.65(n + 1)th term

n = sample size = 13

0.65(14) th term

= 9.1th term

= (9 + 10)th / 2

= (121 + 124) / 2)

= 122.5

(6) What is the IQR?

Q3 - Q1

124.5 - 112

= 12.5

(7) What are the lower and upper limits?

Maximum = 138

Minimum = 65

4 0

2 years ago

A recipe uses 1 cup of sugar to make 6 cookies. If you want to make 30 cookies, how much sugar will you need? Use cups as your l

sergij07 [2.7K]

Answer:

the answer is 5 cups

Step-by-step explanation:

let x rep a cup of sugar

and y rep a cookie

and k be constant

x/kx=6y/30y

k=5

8 0

2 years ago

An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of mov- ing violatio

olga2289 [7]

Answer:

a) $E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

b) $E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

Step-by-step explanation:

We assume the following distribution given:

Y 0 1 2 3

P(Y) 0.60 0.25 0.10 0.05

Part a

We can find the expected value with this formula:

$E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

Part b

If we want to find the expected value of $100 Y^2$ we need to find the expected value of Y^2 and we have:

$E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i)$

And replacing we got:

$E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

4 0

3 years ago