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lianna [129]
3 years ago
6

Rivika used the last four digits of her telephone number,1048, followed by its prime factorization to create her email password

Determine her password
Mathematics
2 answers:
erastova [34]3 years ago
5 0

Answer:

Her password is 1048222131.

Like Angie said, 1048 * 2* 2 * 2 * 131.

But before that you need to decompose 1048  into its prime factors.

The Prime Factorization is:

2 x 2 x 2 x 131

In Exponential Form:

23 x 1311

CSV Format:

2, 2, 2, 131

All Factors of 1048:

1, 2, 4, 8, 131, 262, 524, 1048

Prime Factors Tree

 1048      

 /  \      

2   524    

   /  \    

 2   262  

     /  \  

   2   131

There you go!  If you need more help just reach out to me!

oksano4ka [1.4K]3 years ago
4 0

Answer: I'm pretty sure I can help, just give me like 3 minutes..

-Angie

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A trader made a profit of 24% by selling an article for #372 how much should he have sold it to make a profit of 48%
Ierofanga [76]

Answer: #444

Step-by-step explanation:

First find the original selling price of the article.

When sold for #372 the profit was 24%. Assume the original price is x.

372 = x + 24% * x

372 = 1.24x

x = 372 / 1.24

= #300

If he wanted to make a profit of 48% he should have sold for:

= Original price + (original price * 48%)

= 300 + (300 * 48%)

= #444

4 0
3 years ago
Weight gain during pregnancy. In 2004, the state of North Carolina released to the public a large data set containing informatio
pychu [463]

Answer:

1. B. H0: μ1−μ2=0, HA: μ1−μ2≠0

2. z=1.2114

3. P-value=0.2257

4. Do not reject H0

Step-by-step explanation:

We have to perfomr an hypothesis test to see if there is strong evidence that there is a significant difference between the two population means.

The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2\neq0

Being μ1 the mean average gain for younger mothers and μ2 the mean average gain for mature mothers.

(NOTE: we are comparing means, not proportions, as it is the random variable is the weight gain).

As we are claiming "strong evidence", the level of significance will be 0.01.

For younger mothers, the sample size is n1=840, the sample mean is 30.7 and  the sample standard deviation is s1=14.91.

For mature mothers, the sample size is n2=132, the sample mean is 29.15 and the sample standard deviation is s2=13.46.

The difference between means is

M_d=\mu_1-\mu_2=30.7-29.15=1.55

The standard error of the difference between means is

s_M=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}=\sqrt{\dfrac{14.91^2}{840}+\dfrac{13.46^2}{132}}=\sqrt{ 0.2647+1.3725}=\sqrt{1.6372}\\\\\\s_M=1.2795

Then, the statistic can be calculated as:

z=\dfrac{M_d-(\mu_1-\mu_2)}{s_M}=\dfrac{1.55-0}{1.2795}=1.2114

The P-value for this z-statistic in a two tailed test is:

P-value=2P(z>1.2114)=0.2257

As the P-value is greater than the significance level, the null hypothesis failed to be rejected.

There is no enough evidence to claim that the real average weight gain differs from mature and youger mothers.

5 0
3 years ago
How many two digit numbers are divisible by either 3 or 5?
agasfer [191]
There are 18 that are divisible by 5. 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95

now find which of those are divisible by 3 (divide the numbers by three and if its a whole number that works for example 30 divided by 3 is 10 so thats one) 


8 0
3 years ago
Read 2 more answers
Statistics :Q6..........
scZoUnD [109]
I’m thinking A. Would be the most suitable answer for this one
7 0
3 years ago
Read 2 more answers
Nine new employees, two of whom are married to each other, are to be assigned nine desks that are lined up in a row. If the assi
meriva

Answer:

The probability is 0.8

Step-by-step explanation:

The key to answering this question is considering the fact that the two married employees be treated as a single unit.

Now what this means is that we would be having 8 desks to assign.

Mathematically, the number of ways to assign 8 desks to 8 employees is equal to 8!

Now, the number of ways the couple can interchange their desks is just 2 ways

Thus, the number of ways to assign desks such that the couple has adjacent desks is 2(8!)

The number of ways to assign desks among all six employees randomly is 9!

Thus, the probability that the couple will have adjacent desks would be ;

2(8!)/9! = 2/9

This means that the probability that the couple have non adjacent desks is 1-2/9 = 7/9 = 0.77778

Which is 0.8 to the nearest tenth of a percent

4 0
3 years ago
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