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lidiya [134]
3 years ago
11

Jimmy works part-time at Walmart earning $11 per hour. He is allowed to work anywhere between 0 to 30

Mathematics
2 answers:
Flura [38]3 years ago
5 0

Answer:  The answers are given below.

Step-by-step explanation:  Given that Jimmy works part-time at Walmart earning $11 per hour. He is allowed to work anywhere between 0 to 30  hours per week.

His only expense is $8 per week for an unlimited bus pass. The following function represents his weekly earnings for h hours :

E(h)=11h-8~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

We are to

(A) evaluate E(10) and interpret the meaning in context of the problem.

(B) evaluate E(15) and interpret the meaning in context of the problem.

(C) interpret the statement E(10) < E(15) .

(A) The value of E(10), can be calculated using equation (i), as follows :

E(10)=11\times10-8=110-8=102.

This means Jimmy earns $102 if he works for 10 hours per week.

(B) The value of E(15), can be calculated using equation (i), as follows :

E(15)=11\times15-8=165-8=157.

This means Jimmy earns $157 if he works for 15 hours per week.

(C) We can see that E(10) < E(15).

This means that Jimmy earns less if he works for 10 hours per week than that he earns if we working 15 hours per week.

Thus, all the three parts are answered.

Neko [114]3 years ago
4 0

A because I did this in school

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9*10(90) - 7*3(21) = 69
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Blood is a buffer solution. When carbon dioxide is absorbed into the bloodstream, it produces carbonic acid and lowers the pH. T
Irina18 [472]

Answer:

x ≅ 20.10 torr

Step-by-step explanation:

Given that the equation which models blood pH in the question is;

pH(x)=6.1+log(800/x)

where;

pH = 7.7

x = partial pressure of carbon dioxide in arterial blood, measured in torr.

we are asked to find (x)

In order to do that, we use the given equation:

pH(x)=6.1+log(800/x)

since pH = 7.7

7.7 = 6.1 + log (800/x)

7.7 - 6.1 =  log (800/x)

1.6 =  log (800/x)

10^{1.6}= \frac{800}{x}

x= \frac{800}{10^{1.6}}

x = 20.09509145

x ≅ 20.10 torr

4 0
3 years ago
8.
lukranit [14]

Answer:

4

Step-by-step explanation:

The y-coordinate of the point where both lines intersect will give us the same solution for x.

At y = 4, both equations will have a corresponding x-value of 1.

Therefore, the value of y that that will make both equations have the same solution of x is 4.

7 0
3 years ago
An object is launched from a launching pad 208 ft. above the ground at a velocity of 192ft/sec. what is the maximum height reach
Alexeev081 [22]

Answer:

h(x) = -16x² + 192x + 208

784ft

6 sec

13 sec

Step-by-step explanation:

a)

h(x) = -16x² +vx + h_{o}

here v represent velocity

         h_{o} represent initial height of launch

       

h(x) = -16x² + 192x + 208

b)

h(x) = -16x² + 192x + 208

here a = -16

        b = 192

        c = 208

x = -b/2a

  = -192/2(-16)

  = 6

plug this value in the equation

h(x) = -16(6)² + 192(6) + 208

      = 784ft

e)

Plug h(x)=0 in the equation

0 = -16x² + 192x + 208

divide equation by -16

x² - 12x - 13 = 0

Factors

1x * -13x = -13

1x - 13x = -12

Factorised form

x² - 12x - 13 = 0

x² + x - 13x - 13 = 0

x(x+1) -13(x+1) = 0

(x+1)(x-13) = 0

x = -1

x = 13

Since time can not be negative so we will reject x = -1  

3 0
4 years ago
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
beks73 [17]

Answer:

(5.4582 ; 6.8618)

Step-by-step explanation:

Given the data:

6 10 2 6 3 3 3 6 6 6 6 5 8 9 10 10 7 9 3 6 5 10 9 9 10 3 8 6 6 3 3 6 6 5 4 10 9 3 5 7 10 6 3 8 6 8 3 3 5 5

Sample mean, xbar = Σx / n

n = sample size = 50

ΣX = 308

xbar = 308 / 50 = 6.16

Using a Calculator :

The sample standard deviation, s = 2.469

Confidence interval = xbar ± margin of error

Margin of Error = Tcritical * s/sqrt(n)

Tcritical at 95% ; df = 50 - 1 = 49

Tcritical = 2.010

Hence,

Margin of Error= 2.010 * (2.469/sqrt(50)) = 0.7018

Lower boundary : (6.16 - 0.7018) = 5.4582

Upper boundary : (6.16 + 0.7018) = 6.8618

(5.4582 ; 6.8618)

4 0
3 years ago
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