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Ivahew [28]
3 years ago
7

The original price of a jacket is $70. The sale price is 30% off the original price. What is the amount off the original price (

discount)? How much will you pay for the jacket with the discount?
Mathematics
2 answers:
Mama L [17]3 years ago
3 0

Answer:

$49

Step-by-step explanation:

$70 - 30% = $49

Anarel [89]3 years ago
3 0

Answer

its 47$

Step-by-step explanation:

You divide the number by the percentage which in this problem would look like 0.30 and then you move the decimal to the left once after you get your answer this is about

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Based on the information provided, to feed the second cat it is necessary 1 1/2 extra cans or 1.5 cans.

<h3>How much extra food is needed to feed the second cat?</h3>

This difference can be found by substracting the total cans fed with only one cat to the total fed with two cats:

  • 3 1/2 cans - 2 cans = 1 1/2 cans, which can also be expressed as 1.5 cans.

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What is the result when the number 70 is increased by 10%​
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3 years ago
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Statistics In the manual “How to Have a Number One the Easy Way,” it is stated that a song “must be no longer than three minutes
dimaraw [331]

Answer:

We conclude that the sample is from a population of songs with a mean greater than 210 seconds.

Step-by-step explanation:

We are given that a simple random sample of 40 current hit songs results in a mean length of 252.5 seconds.

Assume that the standard deviation of song lengths is 54.5 sec.

Let \mu = <u><em>population mean length of the songs</em></u>

So, Null Hypothesis, H_0 : \mu \leq 210 seconds      {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}

Alternate Hypothesis, H_A : \mu > 210 seconds      {means that the sample is from a population of songs with a mean greater than 210 seconds}

The test statistics that will be used here is <u>One-sample z-test statistics </u>because we know about population standard deviation;

                              T.S.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean length of songs = 252.5 seconds

            \sigma = population standard deviation = 54.5 seconds

            n = sample of current hit songs = 40

So, <u><em>the test statistics</em></u> =  \frac{252.5-210}{\frac{54.5}{\sqrt{40} } }

                                   =  4.932

The value of z-test statistics is 4.932.

Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.

4 0
3 years ago
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