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Y_Kistochka [10]
3 years ago
10

Karen wrote a six digit number that only has one 8 in it . Kim also wrote a six digit number. The 8 in karens number is worth 10

times as much as the 8 in kims number.
Mathematics
1 answer:
Dahasolnce [82]3 years ago
3 0

the 8 could be tens spot, Hundreds spot, Thousands ... till the end for karens spot

Step-by-step explanation:

Sorry if im wrong.

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it takes 18 3/8 inches of wood to make a frame for a small picture if Ms Jones has 108 inches of wood how many frames can she ma
scZoUnD [109]
It would make 5 complete frames, and about 8/10 of another frame.
108 / 18.375 = 5.8&*)&^&%*$*
I used a calculator.
You could keep adding 18 and 3/8 over and over again until you get right under 108 if you're good at math like me, but most people would have to add 18 and 3/8 separately.

Thanks, and if you liked my answer, please consider giving me the brainliest answer, because I need 5 of them to get to the next level.
6 0
4 years ago
Read 2 more answers
The first quartile of a data set is 18, the median is 25, the third quartile is 28,
laila [671]
The outlier would be shown as a dot so the answer is B. 45
The interquartile range is 28 - 18 = 10
A simple rule is that points more than 1.5xIRQ above or below the medians is an outlier and is shown as a point.
28 + 1.5x10 = 43. 45 is above this value.
7 0
2 years ago
F(x)=3x when x= -2,0,5
ololo11 [35]

Answer:

F(x)=3x when x= -2,0,5

<h2>15</h2>
6 0
3 years ago
19/92 written as a percentage
Solnce55 [7]

Answer:

20.65%

Step-by-step explanation:

\frac{19}{92}   \\  \\ =  \frac{19 \times 100}{92}  \% \\  \\  =   \frac{1900}{92} \% \\  \\  = 20.6521739 \%\\  \\  = 20.65\%

4 0
3 years ago
Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test
BaLLatris [955]

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\  \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}

- Applying the formula, we have:

\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\  \\ t=\frac{5}{3.1752} \\  \\ t\approx1.5747 \end{gathered}

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

7 0
2 years ago
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