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1 year ago

What is the last digit of the product of all the numbers between 11 and 29?

Mathematics

1 answer:

tekilochka [14]1 year ago

5 0

we can conclude that the last digit of the product of all the numbers between 11 and 29 is 0.

<h3>What is the last digit of the product of all the numbers between 11 and 29?</h3>

Here we want to find the last digit of the product between all the whole numbers larger than 11 and smaller than 29.

Then we have the product:

P = 12*13*14*15*16*17*18*19*20*21*22*23*24*25*26*27*28

Now, notice that there is a 20 there.

Any number times 20 will end with a zero, then:

P = 20*(12*13*14*15*16*17*18*19*21*22*23*24*25*26*27*28)

Only with that, we can conclude that the last digit of the product of all the numbers between 11 and 29 is 0.

If you want to learn more about products:

brainly.com/question/10873737

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Dustin and Melanie are playing a game, where two standard, six-sided number cubes are rolled, and the sum of their outcome is fo

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Answer:

Melanie.

Step-by-step explanation:

The possible ways of getting a 6 are (3,3), (2,4), (4,2), (1,5) and (5,1).

That is 5 ways.

The possible ways of getting a 7 are (1,6), (6,1), (5,2),( 2,5), (3, 4), (4,3).

That is 6 ways.

So Melanie made the better decision.

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3 years ago

Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyz

GaryK [48]

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender Mean Std. dev. n

Female 2577.75 1916.29 67

Male 3809.42 2379.47 33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

$M_d=M_1-M_2=2577.75-3809.42=-1231.67$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795$

The t-value for a 95% confidence interval is t=1.96.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.96 \cdot 475.795=932.54$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13$

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

6 0

3 years ago

Two different types of polishing solutions are being evaluated for possible use in a tumble-polish operation for manufacturing i

Nina [5.8K]

Answer:

$z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358$

$p_v =2*P(Z>5.358) = 4.2x10^{-8}$

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.

Step-by-step explanation:

Data given and notation

$X_{1}=253$ represent the number with no defects in sample 1

$X_{2}=196$ represent the number with no defects in sample 1

$n_{1}=300$ sample 1

$n_{2}=300$ sample 2

$p_{1}=\frac{253}{300}=0.843$ represent the proportion of number with no defects in sample 1

$p_{2}=\frac{196}{300}=0.653$ represent the proportion of number with no defects in sample 2

z would represent the statistic (variable of interest)

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

$\alpha=0.05$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference in the the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} - p_2}=0$

Alternative hypothesis: $p_{1} - p_{2} \neq 0$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{253+196}{300+300}=0.748$

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358$

Statistical decision

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

$p_v =2*P(Z>5.358) = 4.2x10^{-8}$

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3 years ago

Ron has two bags of different stickers. The first bag has 3 cars, 2 trucks, and 5 dinosaurs. The second bag has 6 cars, 4 trucks

Katen [24]

1/2 chance to get a dinosaurs sticker for the first bag. 1/6 chance to get a dinosaurs sticker for the second bag.

4 0

3 years ago

Pleaseeee answer correctly !!!!!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!!!

aev [14]

Answer:

132 degrees

Step-by-step explanation:

its the same as the other side since its a rhombus :)

did i help?

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3 years ago