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2 years ago

Paolo paid $28 for a hat that was originally priced at $35. By what percent was the hat discounted?

Mathematics

1 answer:

svetoff [14.1K]2 years ago

8 0

28*100=2800
2800÷35=80
80%

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If 40 percent of a movie ticket costs $5.00 what is 20 percent of the cost of two tickets

Nikolay [14]

Find cost of a ticket:

40% = $5.00

1% = $5 ÷ 40 = 0.125

100% = 0.125 x 100 = $12.50

The cost of a ticket is $12.50

Find 20% of two tickets:

Two tickets cost $12.50 x 2 = $25

20% of $25 = 0.25 x $25 = $5

Answer: 20% of two tickets cost $5

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2 years ago

A tin of paint covers a surface area of 50m2. Each tin costs £5.60. The entire surface of a solid cylindrical rod with diameter

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2 years ago

The center of a hyperbola is (−2,4) , and one vertex is (−2,7) . The slope of one of the asymptotes is 1/2 .

maxonik [38]

Answer:

<h2>The equation is $\frac{(y - 4)^{2} }{9 } - \frac{(x + 2)^{2} }{36 } = 1$ .</h2>

Step-by-step explanation:

The equation of a hyperbola is represented by $\frac{(y - k)^{2} }{b^{2} } - \frac{(x - h)^{2} }{a^{2} } = 1$ , where (h, k) is the center of the hyperbola.

As per the given condition, h = -2 and k = 4.

Thus, the equation becomes $\frac{(y - 4)^{2} }{b^{2} } - \frac{(x + 2)^{2} }{a^{2} } = 1$ .

One vertex is (-2, 7).

Hence, putting x = -2 and y = 7, we get $b^{2} = 9$ .

The slope of the asymptote is $\frac{b}{a}$ .

Hence, $\frac{b^{2} }{a^{2} } = \frac{1}{4} \\a^{2} = 36$ .

Thus, the equation is $\frac{(y - 4)^{2} }{9 } - \frac{(x + 2)^{2} }{36 } = 1$ .

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2 years ago

Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

 $P(D)= \frac{1}{3}$ 

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

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2 years ago

Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to

RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

5 0

2 years ago