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Troyanec [42]
3 years ago
9

griffin ordered a pair of sneakers online. he had 16 credit that he applied toward the purchase, and then he used a credit card

to pay for the rest of the cost. if the the shoes cost 80, then how much did griffin charge to his credit card when he bought the sneakers? PLEASE ANSWER I BEG Y'ALL
Mathematics
1 answer:
Doss [256]3 years ago
4 0

Answer: Griffin charge $79.854 to his credit card when he bought the sneakers.

Step-by-step explanation:

Griffin ordered a pair of sneakers online.

Value of each credit point = 1 cent

Then , value of 16 credit points = 16 cents = $0.16 [1$ = 100 cents]

Cost of shoes = Rs $80

Charge to credit card = (Cost of shoes) - (Value of 16 credit points)

= $(80-0.16)

= $79.84

Hence, Griffin charge $79.854 to his credit card when he bought the sneakers.

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Answer:

Range: 0.07

Median: 0.145

Mode: none

Step-by-step explanation:

<u>Range:</u> largest #- smallest #

<u>Median:</u> middle number (when numbers are in order least to greatest) or when two are middle number you add them, and divide by two

<u>Mode:</u> most occurring number(s) (if there aren't any, no mode)—but if there are multiples such as 2 zeros and 2 ones then 0 and 1 would be the mode

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Ann took a taxi home from the airport. The taxi fare was $2.10 per mile, and she gave the driver a tip of $5. Ann paid a total o
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<span>d = miles from the airport to home. 
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3 years ago
How do I solve for y in the equation 2(3/4)+y=4?
yuradex [85]
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2 years ago
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Create an equivalent trinomial for 2 (3x + 5)(2x - 6)​
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5 0
3 years ago
A local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. In o
Mandarinka [93]

Answer:

(a) Null Hypothesis, H_0 : \mu = 22 ounces  

    Alternate Hypothesis, H_A : \mu\neq 22 ounces

(b) The value of the test statistic is -2.687.

(c) The critical values are -1.96 and 1.96.

(d) We conclude that Reject H_0 since the value of the test statistic is less than the negative critical value.

Step-by-step explanation:

We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.

He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.

<em>Let </em>\mu<em> = average ounces of passion fruit juice used to fill each bottle.</em>

(a) So, Null Hypothesis, H_0 : \mu = 22 ounces     {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}

Alternate Hypothesis, H_A : \mu\neq 22 ounces     {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

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

where, \bar X = sample mean weight of the passion fruit juice = 21.54 ounces

            \sigma = population standard deviation = 1.38 ounce

            n = sample of bottles = 65

So, <u><em>test statistics</em></u>  =  \frac{21.54-22}{\frac{1.38}{\sqrt{65} } }  

                               =  -2.687

(b) The value of z test statistics is -2.687.

(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

<em>Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that Reject H_0 since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.

6 0
3 years ago
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