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djyliett [7]
3 years ago
11

Find the greatest number of six digit which is a perfect square.

Mathematics
2 answers:
Vesna [10]3 years ago
7 0
A "perfect square" is an integer that is the square of an integer.

The largest 6-digit integer is 999,999.

The square root of it is 999.9995

So the greatest square of an integer is the square of 999 = <u>998,001</u> .
Mashutka [201]3 years ago
5 0
From the ques, it can be said that the answer would be less than 1000000
Start from 999,
999*999= 998001
Therefore, 998001 is a greatest number of square 999.
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For population parameter μ = true average resonance frequency for all tennis rackets of a certain type, each of these is a confi
Andreyy89

Answer:

X[bar]= 115

Step-by-step explanation:

Hello!

Every Confidence interval to estimate the population mean are constructed following the structure:

"Estimator" ± margin of error"

Wich means that the intervals are centered around the sample mean. To know the value of the sample mean you have to make the following calculation:

X[bar]= \frac{Upper bond + Low bond}{2}

X[bar]= \frac{115.6+114.4}{2} = 115

Since both intervals were calculated with the information of the same sample, you can choose either to calculate the sample mean.

I hope it helps!

5 0
3 years ago
If I had seven stones and I threw 8 how many stones do I have left??
igomit [66]
It is impossible as 8 is bigger than 7 and you can’t take it away ur welcome
7 0
3 years ago
Read 2 more answers
6/11 of students are at a school are girls . If there are 1309 students at the school , how many are boys
Alexandra [31]
There are 714 boys in that school, you can cross multiply and divide to find the answer, if you wanna check. The way I wrote it was x (boys) / 1309 = 6/11.
3 0
3 years ago
Select the correct answer.
GREYUIT [131]
I don’t know
Igditditditd
4 0
2 years ago
A average amount of money on deposit in a savings account is $7500. Suppose a random sample of 49 accounts shows the average in
Valentin [98]

Answer:  

z=-3.79

p_v =2*P(z  

Step-by-step explanation:  

1) Data given and notation  

\bar X=6850 represent the mean average amount of money on deposit in savings account for the sample  

\sigma=1200 represent the population standard deviation for the sample  

n=49 sample size  

\mu_o =7500 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean for the amount of money on deposit in savings account differs from 7500, the system of hypothesis would be:  

Null hypothesis:\mu =7500  

Alternative hypothesis:\mu \neq 7500  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

z=\frac{6850-7500}{\frac{1200}{\sqrt{49}}}=-3.79  

4)P-value  

Since is a two-sided test the p value would be:  

p_v =2*P(z  

5) Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average amount of money deposit in savings account its significantly different from 7500 tons at 1% of signficance.  

7 0
3 years ago
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