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max2010maxim [7]

3 years ago

5

Jenna earns $8.75 per hour working at the mall. If Jenna worked hours last week, how much did she earn before taxes were taken o

ut?

1 answer:

Romashka [77]3 years ago

5 0

If she earns $8.75 for 1 hour, and works for 20 1/5 hours, she'll earn:
8.75 x 20 1/5 = 176.75
Jenna earns $176.75 before tax

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Ms. Wilson purchased a

snow_lady [41]

Answer: 97.7 in

Step-by-step explanation: The perimeter is another name for circumference. Since we know that the formula for circumference is 2(pi)r, but we only want half of that.

Note that 2r is another way of writing the diameter (d).

In this case, we can use the formula d(pi)/2 + d (we’re adding d because d(pi)/2 will only give us the perimeter of the arch.

38(pi)/2 = 59.66

59.66 + 38 = 97.66

Rounded, that would be 97.7 in

5 0

3 years ago

300x+290 factor out the gfe of each expression

marta [7]

Answer: 590x.

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Step-by-step explanation:

3 0

2 years ago

Janelys needed to get her computer fixed. She took it to the repair store. The technician at the store worked on the computer fo

Sav [38]

Answer: x = 45

Step-by-step explanation:

6 0

2 years ago

Consider this algebraic expression: –3x + x + 5 Use the algebra tiles to model the expression.

luda_lava [24]

For this case, we have the following expression:

$-3x + x + 5$

We simplify the expression:

If we add similar terms, taking into account that different signs are subtracted and the sign of the greater one is placed, we have that $-3x + x = -2x$

So, we have to:

$-3x + x + 5 = -2x + 5$

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-3x + x + 5 = -2x + 5

6 0

3 years ago

Read 2 more answers

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

4 0

2 years ago