If 5.7×10^k= 570,000, what is the value of k?

Evaluate the given numerical expression <br> (5+7 to the 3rd power divided by 7 times 7

-BARSIC- [3]

${(5 + 7)}^{3} \div 7 \times 7 \\ {12}^{3} \div 7 \times 7 \\ 1728 \div 7 \times 7 \\ 1728$

8 0

3 years ago

What are all the fractors of 21 and of 30?

iren2701 [21]

21= 1,3,7,21
30= 1,2,3,5,6,10,15,30

common factors= 1,3

8 0

3 years ago

I need answer Immediately pls!!!!!!

Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

$n(U)=27$

$n(A)=7$

$n(B)=18$

$n(A'\cap B')=7$

We know that,

$n(A\cup B)=n(U)-n(A'\cap B')$

$n(A\cup B)=27-7$

$n(A\cup B)=20$

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=7+18-n(A\cap B)$

$n(A\cap B)=7+18-20$

$n(A\cap B)=25-20$

$n(A\cap B)=5$

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

$\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}$

$\text{Probability}=\dfrac{5}{27}$

Therefore, the required probability is $\dfrac{5}{27}$ .

3 0

3 years ago

Each section of a running trail is 3/4 mile. The trail has 3 sections. Which expression tells how many miles long the entire tra

allsm [11]

3/4*3=2.25 miles long trail
Have a great day

8 0

3 years ago

Media experts claim that daily print newspapers are declining because of Internet access. Listed below, from left to right and

yKpoI14uk [10]

Answer:

(a) The median is 1478

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The test statistic $t_{\alpha/2}$ is 6.678155

(d) The p value is 1.2×10⁻¹¹

(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high

Step-by-step explanation:

(a) Here we have the data as follows;

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The median is = 1478

Therefore we have above the median

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484

The mean, $\bar{x}_{1}$ = 1541.9

Standard deviation, σ₁ = 41.38224257

n₁ = 10

Below the median

1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The mean, $\bar{x}_{2}}$ = 1433.9

Standard deviation, σ₂ = 30.04812806

n₂ = 10

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The formula for t test is given by;

$t_{\alpha/2} =\frac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}$

df = 10 - 1 = 9, α = 0.05

Therefore, the test statistic $t_{\alpha/2}$ = 6.678155

(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹

Critical z at 5% confidence level = 1.645

Since P << 0.05, e reject the

(e) Therefore, we reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high.

5 0

3 years ago