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3 years ago

What would the circumference be of a circle if the radius is 3 units

1 answer:

7 0

If we have a circle with a radius of 3 we need to find the diameter of the circle. Remember radius is half of the diameter. So, the diameter is 6 units.

Use the circumference formula: C=3.14D

C=3.14(6)

C=18.84 units

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A restraint bill comes to $28.35. Find the total cost if the tax is 6.25% and a 20% tip is left in the amount before tax.

Fudgin [204]

11 dollar tip and 3.15 dollar tax

3 0

3 years ago

a total of 408 tickets were sold for the school play. They were either adult tickets or student tickets. The number of students

Mama L [17]

<span>Given Situation:
=> There are 408 total of ticket sold for a school play
=> 3 times of the tickets are sold to students compare to the adults
=> So the ratio of the given situation is 1 : 3
=> Now, 1 + 3 = 4
Since, we need 4 parts, let’s divide 408 by 4
=> 408 / 4 = 102 – the number of tickets for adults
=> 102 x 3 = 306 – the number of tickets for students
=> 102 : 306</span>

4 0

3 years ago

Jeff has 8 red marbles 6 blue marbles, and 4 green marbles that are the same size and shape he puts the marbles into a bag, mixe

Cloud [144]

Hey buddy I am here to help!

Red = 8

blue = 6

greem = 4

total = 18

probability of blue marble = 6/18 = 1/3

4 0

3 years ago

Read 2 more answers

At Cheng's Bike Rentals, it costs $25 to rent a bike for 5 hours.

timurjin [86]

Answer:

25÷5=5

So the answer is (5)

7 0

2 years ago

A new sales training program has been instituted at a rent-to-own company. Prior to the training, 10 employees were tested on th

Lunna [17]

Answer:

C. H0: µD = 0, HA: µD <0

Step-by-step explanation:

We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value for pretest , y = test value for posttest

x: 66, 94, 87, 84, 76, 88

y: 75, 100, 93, 85, 75, 90

The system of hypothesis for this case are:

Null hypothesis: $\mu_D \geq 0$

Alternative hypothesis: $\mu_D < 0$

Because the difference D is defined as Pretest-Postest. And we want to see if the postets score is higher than the pretest with th training.

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: -9, -6, -6, -1, 1, -2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-23}{6}=-3.833$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.763$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-3.833 -0}{\frac{3.763}{\sqrt{6}}}=-2.494$

The next step is calculate the degrees of freedom given by:

$df=n-1=6-1=5$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v =P(t_{(5)}$

The p value is lower than the significance level assumed $\alpha=0.05$ , so then we can conclude that we can reject the null hypothesis. So we can say that the differences between Pretest and Posttest $\mu_{pretest}-\mu_{postest}$ are less than 0.

3 0

3 years ago