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A total of 26 bills are in a cash box. Some of the bills are one-dollar bills, and the rest are five-dollar bills. The total amo

kobusy [5.1K]

Answer:

Number of 1 $ bill = 20

Number of 5 $ bill = 6

Explanation:

Let p represent number of one dollar bolls and q represent number of five bills.

A total of 26 bills are in a cash box.

p + q = 26 ------------------------- Equation 1

The total amount of cash in the box is $50

p + 5q = 50 ------------------------- Equation 2

Equation 2 - Equation 1,

p + 5q - p -q = 50 - 26

4q = 24

q = 6

Substituting in equation 1.

p + 6 = 26

p = 20

Number of 1 $ bill = 20

Number of 5 $ bill = 6

3 0

3 years ago

The values of 60, 62, and 84 were common to both samples. The three values are identified as outliers with respect to the age-gr

postnew [5]

Answer:

Only 60 is an identified outlier

Step-by-step explanation:

Given

Age group: 40 to 50

$Min = 60$ $Q_1 = 70$ $Median= 73$

$Q_3 = 76$ $Max = 85$

Conditions for outlier

$Value > Q_3 + 1.5 * IQR$

$Value < Q_1 - 1.5 * IQR$

See attachment for complete question

Required

Which of 60, 62 and 84 is an outlier of age grout 40 to 50 years

First, calculate the IQR of group 40 to 50.

This is calculated as:

$IQR = Q_3 - Q_1$

Where:

$Q_3 = 76$ and $Q_1 = 70$

So:

$IQR = 76 - 70$

$IQR = 6$

Next, is to test the outlier conditions on each value (i.e. 60, 62 and 84)

Testing 60

Condition 1

$Value > Q_3 + 1.5 * IQR$

$60 > 76 + 1.5 * 6$

$60 > 76 + 9$

$60 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$60 < 70 - 1.5 * 6$

$60 < 70 - 9$

$60 < 61$ --- True

Because one of the conditions is true, then 60 is an outlier of group 40 - 50 years

Testing 62

Condition 1

$Value > Q_3 + 1.5 * IQR$

$62> 76 + 1.5 * 6$

$62 > 76 + 9$

$62 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$62 < 70 - 1.5 * 6$

$62 < 70 - 9$

$62 < 61$ --- False

Because both conditions are false, then 62 is not an outlier of group 40 - 50 years

Testing 84

Condition 1

$Value > Q_3 + 1.5 * IQR$

$84> 76 + 1.5 * 6$

$84 > 76 + 9$

$84 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$84 < 70 - 1.5 * 6$

$84 < 70 - 9$

$84 < 61$ --- False

Because both conditions are false, then 84 is not an outlier of group 40 - 50 years

5 0

2 years ago

What is the value of the function at x=−2?

adell [148]

x = -2

function value at x = -2 is intersection of the x value and graph

which is, 3

Hope it is clear:)

5 0

2 years ago

Solve. 8x² + 5 = 35 Round to the nearest hundredth. Enter your answers in the boxes. The solutions are approximately and .

lora16 [44]

Answer:

x=1.94

x = - 1.94

Step-by-step explanation:

8x² + 5 = 35

Subtract 5 from each side

8x² + 5-5 = 35-5

8x² = 30

Divide each side by 8

8x² /8 = 30/8

x² = 15/4

Take the square root of each side

sqrt( x²) = ±sqrt(15/4)

x = ±sqrt(15/4)

x=1.93649

x = - 1.93649

To the nearest hundredth

x=1.94

x = - 1.94

4 0

2 years ago

Read 2 more answers

Consider the line y=9x-5.

igomit [66]

Y = 9x - 5...the slope here is 9...a parallel line will have the same slope.

y = mx + b
slope(m) = 9
(5,-4)...x = 5 and y = -4
now we sub and solve for b, the y int
-4 = 9(5) + b
-4 = 45 + b
-4 - 45 = b
-49 = b
so ur parallel line is : y = 9x - 49

y = 9x - 5...slope is 9. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So the slope we need is -1/9.

y = mx + b
slope(m) = -1/9
(5,-4)...x = 5 and y = -4
now we sub and solve for b, the y int
-4 = -1/9(5) + b
-4 = -5/9 + b
-4 + 5/9 = b
-36/9 + 5/9 = b
- 31/9 = b

so ur perpendicular line is : y = -1/9x - 31/9

6 0

2 years ago