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3 years ago

9

a box or laundry detergent is on sale for $6.50. If this price represents a 40% discount from the original price, what is the or

iginal price to the nearest cent?

2 answers:

Troyanec [42]3 years ago

8 0

To solve the question, you divide 6.50 by 60%, or 0.60

the answer to that is $10.83

Usimov [2.4K]3 years ago

4 0

The discount is $2.60. $6.50 + $2.60 = $9.10. $9.10 is the original price.

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Over the last three evenings, Debra received a total of 110 phone calls at the call center. The second evening, she received 4 t

Elza [17]

Answer:

See below.

Step-by-step explanation:

Total = 110

Let calls on third evening = x

second evening = 4x

first evening = x + 8

=> x + 4x + x + 8 = 110

=> 6x = 102

=> x = 17

First evening = x = 17 calls

Second evening = 4x = 68 calls

Third evening = x + 8 = 25 calls

4 0

1 year ago

A line is defined by the equation y = 2/3 x - 6 The line passes through a point whose y-coordinate is 0. What is the x-coordinat

Iteru [2.4K]

Answer:

x = 9

Step-by-step explanation:

Use the equation of the line, and let y = 0. Then solve for x.

y = 2/3 x - 6

Let y = 0:

0 = 2/3 x - 6

Add 6 to both sides.

6 = 2/3 x

Multiply both sides by 3/2.

3/2 * 6 = x

x = 9

8 0

3 years ago

2. The actual tree on the left is 10 feet tall. The model tree on the right is 8 inches tall. What is the scale factor?

Lina20 [59]

Answer:

D. 1/15

Step-by-step explanation:

Actual tree = 10 ft = 12 × 10 = 120 in.

Model tree = 8 in.

Scale factor = model tree/actual tree

Scale factor = 8/120

Scale factor = 1/15

4 0

2 years ago

If value of a+b=7and ab=12then find the value of a and b

LiRa [457]

a+b = 7

ab =12

we know that

a and b = ab – a+b

= 12 –7

= 5 ans

3 0

2 years ago

Read 2 more answers

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago