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tankabanditka [31]
1 year ago
10

Estimation Gabe goes to the mall. If k is the number of items he bought, the exp11.02k +21 gives the amount he spent in dollars

at one store. Then he spent 30 dollars at anotherstore. Find the expression which represents the amount Gabo spent at the mall. Then estimatehow much Gabe spent if he bought 2 items.What expression represents the amount in dollars that Gabe spent at the mall?
Mathematics
1 answer:
Lelechka [254]1 year ago
6 0

The expression 11.02k+21 indicates how much did Gabe spend in the first store he visited.

In the second store, he spent 30 dollars.

To determine how much he spent in the mall you have to add the amounts he spent in both stores:

\begin{gathered} \text{Store}1+\text{Store}2 \\ (11.02k+21)+30 \\ 11.02k+21+30 \\ 11.02k+51 \end{gathered}

The expression that represents his total expenses at the mall is

11.02k+51

"k" represents the number of items he bought. If he bought k=2 items, you can calculate how much he spent at the mall as follows:

\begin{gathered} 11.02k+51 \\ 11.02\cdot2+51 \\ 72.04 \end{gathered}

Gabe spent an estimate of 72.04 dollars at the mall

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Step-by-step explanation:

The probability of an events = P= 2/7 = 0.2857

The probability of its compliments = 1-p = 1-0.2857 = 0.7143 = 5/7

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A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with σ=0.0
oksian1 [2.3K]
<h2>Answer with explanation:</h2>

The confidence interval for population mean is given by :-

\overline{x}\pm z^* SE                   (1)

, where \overline{x} = sample mean

z* = critical value.

SE = standard error

and SE=\dfrac{\sigma}{\sqrt{n}} , \sigma = population standard deviation.

n= sample size.

As per given , we have

\overline{x}=74.021

\sigma=0.001

n= 15

It is known that ring diameter is normally distributed.

SE=\dfrac{0.001}{\sqrt{15}}=0.000258198889747\approx0.000258199

By z-table ,

The critical value for 95% confidence  = z*= 1.96

A 99% two-sided confidence interval on the true mean piston diameter :

74.021\pm (2.576) (0.000258199)     (using (1))

74.021\pm 0.000665120624

74.021\pm 0.000665120624\\\\=(74.021- 0.000665120624,\ 74.021+ 0.000665120624)\\\\=(74.0203348794,\ 74.0216651206)\approx(74.020,\ 74.022)  [Rounded to three decimal places]

∴  A 99% two-sided confidence interval on the true mean piston diameter  = (74.020, 74.022)

By z-table ,

The critical value for 95% confidence  = z*= 1.96

A 95% lower confidence bound on the true mean piston diameter:

74.021- (1.96) (0.000258199)    (using (1))

74.021- 0.00050607004=74.02049393\approx74.020 [Rounded to three decimal places]

∴  A 95% lower confidence bound on the true mean piston diameter= 74.020

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