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3 years ago

How many prime numbers are between 1 and 100

Mathematics

2 answers:

sweet [91]3 years ago

8 0

There are 25 prime numbers.

leonid [27]3 years ago

8 0

There are 25 prime numbers between 1 and 100, with them being:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.
Hope this helps :-)

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A copyeditor thinks the standard deviation for the number of pages in a romance novel is six. A sample of 25 novels has a standa

alina1380 [7]

Answer:

No, the standard deviation for number of pages in a romance novel is six only.

Step-by-step explanation:

First we state our Null Hypothesis, $H_o$ : $\sigma$ = 6

and Alternate Hypothesis, $H_1$ : $\sigma$ > 6

We have taken these hypothesis because we have to check whether our population standard deviation is higher than what editor hypothesized of 6 pages in a romance novel.

Now given sample standard deviation, s = 9 and sample size, n = 25

To test this we use Test Statistics = $\frac{(n-1)s^{2} }{\sigma^{2} }$ follows chi-square with (n-1) degree of freedom [ $\chi ^{2}_n__-1$ ]

Test Statistics = $\frac{(25-1)9^{2} }{6^{2} }$ follows $\chi ^{2}_2_4$ = 54

and since the level of significance is not stated in question so we assume it to be 5%.

Now Using chi-square table we observe at 5% level of significance the $\chi ^{2}_2_4$ will give value of 36.42 which means if our test statistics will fall below 36.42 we will reject null hypothesis.

Since our Test statistics is more than the critical value i.e.(54>36.42) so we have sufficient evidence to accept null hypothesis and conclude that our population standard deviation is not more than 6 pages which the editor hypothesized.

8 0

3 years ago

Write the expression as a cube of monomial: 27x^12

Gnoma [55]

Step-by-step explanation:

ax=axn to rewrite

3√27x12 as (27x12)1/3

to rewrite

3√(3x^4)3

Pull terms out from under the radical, assuming real numbers.

3X^4

3 0

3 years ago

Breanna thought of a number, added 6 to it, multiplied it by 6, subtracted 6 from her answer, and then divided it by 6. She got

ss7ja [257]

Answer:5

Step-by-step explanation:

If you look at the question in reverse you can backtrack your way through. Take 10 and instead of dividing, multiply it by 10, which is 60, add 6 instead of subtracting, which is 66, divide by 6 instead of multiplying, which is 11 and finally subtract 6 instead of adding, which is 5. Use five and try the question over to make sure you calculated correct.

8 0

3 years ago

-4 x (2w - 5) x (w+3). can someone help me with this please?

NeTakaya

Answer:-8w^2x^2-4wx^2+60x^2

Step-by-step explanation:

4 0

4 years ago

Students in a representative sample of 69 second-year students selected from a large university in England participated in a stu

Serhud [2]

Answer:

95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

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

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean procrastination score = 41

s = sample standard deviation = 6.89

n = sample of students = 69

$\mu$ = population mean estimate

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

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

P(-1.9973 < $t_6_8$ < 1.9973) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.9973 & 1.9973 with P = 2.5%}

P(-1.9973 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.9973) = 0.95

P( $-1.9973 \times{\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ =[ $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ , $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ]

= [ $41-1.9973 \times{\frac{6.89}{\sqrt{69} } }$ , $41+1.9973 \times{\frac{6.89}{\sqrt{69} } }$ ]

= [39.34 , 42.66]

Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

5 0

4 years ago