Write 7√5 using rational exponents.

How does the value of the digit 3 in the number 63,297 compared to the value of the digit 3 in the number 60,325?

Elina [12.6K]

Answer:

The value of digit 3 in 63,297 is in the thousands place while the digit 3 in the number 60,325 is in the hundreds place

Hope this helps!

4 0

4 years ago

A certain lottery game, 32 numbers are randomly mixed and 7 are selected. A person must pick all 7 numbers to win. Order is not

lorasvet [3.4K]

Answer:

Below.

Step-by-step explanation:

The number of combinations of 7 from 32 = 32! / 7! (32-7)!

= 3365856

So it is 1/ 3365856 = 0.000000297

or about 3 in 10,000,000.

3 0

3 years ago

HELP PLEASE !!!!!!!!!!!!!!!

valina [46]

(f×g)(x) = (fg)(x)

(f×g)=(3x²+5)× (x-2)

use the foil method:

(fg)(x)=3x³-6x²+5x-10

so it is C which is correct

4 0

3 years ago

An article reported on the results of an experiment in which half of the individuals in a group of 66 postmenopausal overweight

kicyunya [14]

Answer:

We conclude that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg.

Step-by-step explanation:

We are given that an article reported on the results of an experiment in which half of the individuals in a group of 66 postmenopausal overweight women were randomly assigned to a particular vegan diet, and the other half received a diet based on National Cholesterol Education Program guidelines.

The sample mean decrease in body weight for those on the vegan diet was 6 kg, and the sample SD was 3.2, whereas, for those on the control diet, the sample mean weight loss and standard deviation were 3.8 and 2.4, respectively.

Let = true average weight loss for the vegan diet.

$\mu_2$ = true average weight loss for the control diet.

So, Null Hypothesis, : 1 kg {means that the true average weight loss for the vegan diet exceeds that for the control diet by less than or equal to 1 kg}

Alternate Hypothesis, : > 1 kg {means that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1-\barX_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, = sample mean weight loss for the vegan diet = 6 kg

= sample mean weight loss for the control diet = 3.8 kg

= sample standard deviation weight loss for the vegan diet = 3.2 kg

= sample standard deviation weight loss for the control diet = 2.4 kg

$n_1$ = sample of vegan diet women = 33

$n_2$ = sample of control diet women = 33

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(33-1)\times 3.2^{2}+(33-1)\times 2.4^{2} }{33+33-2} }$ = 2.83

So, the test statistics = $\frac{(6-3.8)-(1)}{2.83 \times \sqrt{\frac{1}{33}+\frac{1}{33} } }$ ~ $t_6_4$

= 1.722

The value of t-test statistics is 1.722.

Also, the P-value of the test statistics is given by;

P-value = P( $t_6_4$ > 1.722) = 0.0461 or 4.61%

Since the P-value of our test statistics is less than the level of significance as 0.0461 < 0.05, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg.

3 0

3 years ago

If you divide any natural number n by 4, you get a remainder r.

blondinia [14]

Given:

If you divide any natural number n by 4, you get a remainder r.

To find:

The values of r if $n=13;34;43;100$ . Also find the domain and range.