What number has a prime factorization of 2x 5^2

m/6=15/18 I need help also I need to provide work so can somebody explain it or send a picture please? Thank you^^

Lostsunrise [7]

Answer:

m=3

Step-by-step explanation:

Hello, Hope it helps :-)

4 0

3 years ago

A newsletter publisher believes that above 41A% of their readers own a personal computer. Is there sufficient evidence at the 0.

klemol [59]

Complete Question

A newsletter publisher believes that above 41% of their readers own a personal computer. Is there sufficient evidence at the 0.10 level to substantiate the publisher's claim?

State the null and alternative hypotheses for the above scenario.

Answer:

The null hypothesis is $H_o : p \le 0.411$

The alternative hypothesis is $H_a : \mu > 0.41$

Step-by-step explanation:

From the question we are told that

The proportion is $p = 0.41$

Generally the null hypothesis is $H_o : p \le 0.411$

The above condition represents the null hypothesis because it contains and equality in its condition

The alternative hypothesis is $H_a : \mu > 0.41$

6 0

3 years ago

If (-2,3), 3x-2ky+4=0 find the value of k

ipn [44]

Answer: k = -1/3

Step-by-Step Explanation:

=> 3x - 2ky + 4 = 0
Value of ‘x’ = -2
Value of ‘y’ = 3

Substitute values of ‘x’ and ‘y’ :-
=> 3x - 2ky + 4 = 0
= 3(-2) - 2k(3) + 4 = 0
= -6 - 6k + 4 = 0
= -6k = 6 - 4
= -6k = 2
= k = 2/-6
=> k = -2/6 = -1/3

Therefore, k = -1/3

8 0

2 years ago

A researcher is interested in finding a 95% confidence interval for the mean number of times per day that college students text.

Oxana [17]

Complete Question

A researcher is interested in finding a 95% confidence interval for the mean number of times per day that college students text. The study included 210 students who averaged 28 texts per day. The standard deviation was 21 texts.

Round your answers to two decimal places.

A. The sampling distribution follows a [ Select ] ["F", "normal", "T", "Chi-square"] distribution.

B. With 95% confidence the population mean number of texts per day of is between [ Select ] ["24.92", "25.79", "27.37", "25.14"] and [ Select ] ["31.19", "31.20", "29.28", "30.86"] texts.

C. If many groups of 210 randomly selected students are studied, then a different confidence interval would be produced from each group. About [ Select ] ["5", "95", "1", "99"] percent of these confidence intervals will contain the true population mean number of texts per day and about [ Select ] ["95", "99", "5", "1"] percent will not contain the true population mean number of texts per day.

Answer:

a. With 95% confidence the population mean number of texts per day is between $I = 21.06$ and $I = 30.33$ texts.

b

If many groups of 132 randomly selected members are studied, then a different confidence interval would be produced from each group. About <u>95%</u> percent of these confidence intervals will contain the true population number of texts per day and about <u>5%</u> percent will not contain the true population mean number of texts per day.

Step-by-step explanation:

From the question we are told that

The sample size is $n = 102$

The population mean is $\mu = 25.7$

The standard deviation is $\sigma = 23.9$

Generally the standard error of mean is mathematically evaluated as

$\sigma_e = \frac{\sigma}{\sqrt{n } }$

substituting values

$\sigma_e = \frac{23.9}{\sqrt{102 } }$

$\sigma_e = 2.3665$

Given that the level of confidence is 95% then the level of significance will be

$\alpha = 100 -95$

$\alpha =$ 5%

=> $\alpha = 0.05$

Now the critical values of this level of significance is obtained from the critical value table as

$t_{n,\alpha } = t_{102,0.05} = 1.96$

Generally the 95% confidence interval is mathematically represented as

$I = \mu \pm t_{102, 0.05} * \sigma_e$

substituting values

$I = 25.7 \pm 1.96 *2.3665$

$I = 25.7 \pm 4.638$

$I = 30.33$ and $I = 21.06$

So the interval is

6 0

4 years ago

Big Enos loves Coors beer but he lives in Georgia and nobody sells Coors east of Texas. So he offers to pay Cletus "The Snowman"

Alik [6]

Answer:

Snowman Expenses:

a) Income Tax deductible expenses:

Payment for 400 cases of beer = $10,000

Payment for fuel = $650

Total = $10,650

b) The expenses above are allowable because they are "ordinary and necessary expenses incurred in carrying on any trade or business." IRS Code 162.

c) The bribe paid to the Sheriff of $1,000 is not allowable under the same code.

d) Such payments are not allowable because, according to IRS Code 162 section C: 1 and 2, they are described as "Illegal bribes, kickbacks, and other payments" and "other illegal payments."

Step-by-step explanation:

a) According to the Internal Revenue Code 162 which details trade or business expenses, section A specifically state that "There shall be allowed as a deduction all the ordinary and necessary expenses paid or incurred during the taxable year in carrying on any trade or business, including—

(1)a reasonable allowance for salaries or other compensation for personal services actually rendered;

(2)traveling expenses (including amounts expended for meals and lodging other than amounts which are lavish or extravagant under the circumstances) while away from home in the pursuit of a trade or business; and ..."

b) Bribes are illegal payments prohibited by states law. Snowman paid the payment in order not to get his vehicle impounded and himself imprisoned, according to the Sheriff's threats. That he was over-speeding in order to meet his business appointment does not guarantee him the right to bribe a government official.

7 0

4 years ago