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2 years ago

12

A sporting goods store is selling all their exercise equipment for 30% off. The regular price for an exercise bicycle is $260. T

he regular price for a set of hand weights is $120.

1 answer:

katrin [286]2 years ago

3 0

Answer:

Don't know what the question is but if everything's 30% off then the bicycle would now be $182 and the hand weights would be $84.

Step-by-step explanation:

Just multiply the original value by the percentage in decimal form and take the product and subtract that value from the original amount.

260*0.3=78

260-78=182

Hope this helps

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11) The Cost of maintaing a

dezoksy [38]

Answer:

a). Cost of 44 pupils = $14265

b). Least number of pupils = 31

Step-by-step explanation:

The given question is incomplete; here is the complete question.

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.

(a) Find the cost when there are 44 pupils.

(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?

Let the equation representing the total cost of maintaining a school is,

C = ax + b

Where C = Total cost of maintaining a school

a = Fee per pupil

b = Fixed running cost

x = number of pupils

a). Cost of 50 pupils = $15705

Equation will be,

15705 = 50a + b -------(1)

Cost of 40 pupils = $13305

Equation will be,

13305 = 40a + b --------(2)

By subtracting equation (2) from equation (1),

15705 - 13305 = (50a + b) - (40a + b)

2400 = 10a

a = 240

From equation (1),

b = 3705

Equation representing the total cost will be,

C = 240x + 3705

If x = 44

C = 240(44) + 3705

C = $14265

b). If the fee per pupil 'a' = $360

Let the number of pupils = p

Total fee of 'p' pupils = $360p

Total cost to run the school will be = 3705 + 240p

For the school not to be in the loss,

360p ≥ 3705 + 240p

360p - 240p ≥ 3705

120p ≥ 3705

p ≥ $\frac{3705}{120}$

p ≥ 30.875

Therefore, to run the school without loss, number pupils should be at least 31.

3 0

2 years ago

Which expression is equivalent to sin7pi/6?

elena-14-01-66 [18.8K]

For this case we must find an expression equivalent to:

$Sin (\frac {7 \pi} {6})$

According to the unit circle, $\frac {7 \pi} {6}$ is in the third quadrant and equals 210 degrees.

Then, we must find an expression equivalent to:

$Sin (210) = - \frac {1} {2} = - 0.5$

We have to:

$\frac{11 \pi} {6}$

It is in the fourth quadrant and is equivalent to 330 degrees, according to the unit circle.

$Sin (330) = - \frac {1} {2} = - 0.5$

Answer:

Option D

3 0

2 years ago

Read 2 more answers

Jack is driving from his house to the airport. At 9:00 AM, he is 150 miles from the airport. At 10:30 AM, he is 75 miles from th

Jet001 [13]

Answer:

I b3live it is 50 miles/hour

Step-by-step explanation:

We need to know how many miles he drove from 9:00 to 10:30 so you do 150-75 which is 75. Then, you do 75÷1.5 because 9:00 and 10:30 are an hour and a half apart and then you get 50 mph.

4 0

2 years ago

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

Viktor [21]

Answer:

Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

$T(t)=230 -e^{-kt}$

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

$138=230 -Ae^{-k\times 18}$

$\Rightarrow -Ae^{-18k}=138-230$

$\Rightarrow Ae^{-18k}=92$ .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

$184=230-Ae^{-k\times 36}$

$\Rightarrow Ae^{-36k}=230-184$

$\Rightarrow Ae^{-36k}=46$ .......(2)

Dividing (2) by (1)

$\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}$

$\Rightarrow e^{-18k}=\frac{46}{92}$

Taking ln both sides

$ln e^{-18k}=ln\frac{46}{92}$

$\Rightarrow -18k =ln (\frac12)$

$\Rightarrow -18k= ln1-ln2$

$\Rightarrow k= \frac{ln2 }{18}$

Putting the value k in equation (1)

$Ae^{-18\frac{ln2}{18}}=92$

$\Rightarrow A e^{ln2^{-1}}=92$

$\Rightarrow A.2^{-1}=92$

$\Rightarrow \frac{A}{2}=92$

$\Rightarrow A= 92 \times 2$

⇒A= 184.

Therefore $k= \frac{ln2 }{18}$ , A=184

7 0

2 years ago

Suppose the dealer incentive per vehicle for honda's acura brand in 2012 is thought to be bell-shaped and symmetrical with a mea

andreev551 [17]

Answer:

From $1600 to $3400.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 2500

Standard deviation = 300

What interval of dealer incentives would we expect approximately 99.7% of vehicles to fall within?

By the Empirical Rule, 99.7% fall within 3 standard deviations frow the mean. So

From 2500 - 3*300 = 1600 to 2500 + 3*300 = 3400.

8 0

2 years ago