What is the greatest common factor of 4, 37 and 51

I really need help with the whole page. Any help will be really helpful.

Y_Kistochka [10]

The x is going by 15 and the y is going by 5

7 0

3 years ago

Jackrabbits are capable of reaching speeds up to 40 miles per hour. How fast is this in feet per second? (Round to the nearest w

vitfil [10]

58.6667
Underline the 6 and remove the other 2 though

8 0

3 years ago

Read 2 more answers

Write the first six terms of the geometric sequence with the first term 6 and common ratio 1/3

Alja [10]

Answer:

6, 2, 2/3, 2/9, 2/27, 2/81

Step-by-step explanation:

The nth term of a geometric progression is expressed as;

Tn = ar^n-1

a is the first term

n is the number of terms

r is the common ratio

Given

a = 6

r = 1/3

when n = 1

T1 = 6(1/3)^1-1

T1 = 6(1/3)^0

T1 = 6

when n = 2

T2= 6(1/3)^2-1

T2= 6(1/3)^1

T2 = 2

when n = 3

T3 = 6(1/3)^3-1

T3= 6(1/3)^2

T3= 6 * 1/9

T3 = 2/3

when n = 4

T4 = 6(1/3)^4-1

T4= 6(1/3)^3

T4= 6 * 1/27

T4 = 2/9

when n = 5

T5 = 6(1/3)^5-1

T5= 6(1/3)^4

T5= 6 * 1/81

T5 = 2/27

when n = 6

T6 = 6(1/3)^6-1

T6= 6(1/3)^5

T6= 6 * 1/243

T6 = 2/81

Hence the first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81

6 0

3 years ago

On a school trip, the ratio of the teachers to students is 2:21. The ratio boys to girls is 4:3. If there are 18 girls girls on

Alexandra [31]

Answer:

4

Step-by-step explanation:

Let's say B is the number of boys and T is the number of teachers.

T / (B + 18) = 2/21

B/18 = 4/3

Solve the second equation for B.

B = 24

Plug into the first equation to find T.

T / (24 + 18) = 2/21

T = 4

6 0

3 years ago

Conduct the appropriate hypothesis test and compute the test statistic. A company that produces fishing line undergoes random te

andre [41]

Answer:

Option D) Yes, because the test statistic is -2.01

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 pound

Sample mean, $\bar{x}$ = 29.1 pounds

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 2 pounds

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 30\text{ pounds}\\H_A: \mu < 30\text{ pounds}$

We use one-tailed(left) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{29.1 - 30}{\frac{2}{\sqrt{20}} } = -2.012$

Now,

$t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = -1.729$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. Thus, there were enough evidence to conclude that the fishing line breaks with an average force of less than 30 pounds.

Option D) Yes, because the test statistic is -2.01

5 0

4 years ago