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3 years ago

If you had a coupon for 50% of Italian ice and you regularly paid 9.99 for one without discounts how much are you paying?

Mathematics

2 answers:

weeeeeb [17]3 years ago

7 0

If you had a 50% coupon off of $9.99, the end price would be $4.99.

UNO [17]3 years ago

3 0

You would have to divide 9.99 by 2.

so with your coupon you would end up paying 4.99

another way that you can do this is by multiplying 9.99 by .5 and that would give you the same answer

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For the expansion of (x-y)^7 find the indicated term. sixth term

RoseWind [281]

What do you mean by this question?

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3 years ago

The shark are fed three times a day during the morning feeding 2/15 of a ton. During the afternoon feeding the weight of fish fe

UNO [17]

Answer:

The shark are fed $\frac16 \ ton$ of fish during the night.

Step-by-step explanation:

Given:

Weight of fish fed in the morning = $\frac{2}{15}\ ton$

Also Given:

During the afternoon feeding the weight of fish fed will be 1/15 of a ton more than the fish fed during the morning.

Weight of fish fed in the afternoon = $\frac{2}{15}+\frac{1}{15}=\frac{2+1}{15}=\frac{3}{15}\ ton$

Total fish fed in whole day = $\frac12 \ ton$

We need to find the Weight of fish fed in the night.

Solution:

Now we can say that;

Weight of fish fed in the night can be calculated by by subtracting Weight of fish fed in the morning and Weight of fish fed in the afternoon from Total fish fed in whole day .

framing in equation form we get;

Weight of fish fed in the night = $\frac{1}{2}-\frac{2}{15}-\frac{3}{15}= \frac{1}{2}-(\frac{2}{15}+\frac{3}{15})=\frac{1}{2}-\frac{2+3}{15}= \frac{1}{2}-\frac{5}{15} = \frac{1}{2}-\frac{1}{3}$

Now we will use LCM for making the denominator common we get;

Weight of fish fed in the night = $\frac{1\times3}{2\times3}-\frac{1\times2}{3\times2} = \frac{3}{6}-\frac{2}{6}$

Now denominator are common so we will solve the numerators.

Weight of fish fed in the night = $\frac{3-2}{6}=\frac16 \ ton$

Hence The shark are fed $\frac16 \ ton$ of fish during the night.

3 0

3 years ago

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

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3 years ago

Which expression is equivalent to 4.4.4.4.4.4.4.4?

seraphim [82]

Answer:

your answer is D

Step-by-step explanation:

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3 years ago

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