All categories

2 years ago

What is the probability of spinning an odd number on the spinner below?

Mathematics

1 answer:

Ket [755]2 years ago

4 0

If the spinner has an odd number of numbers on it, The probability or spinning an odd number will vary

You might be interested in

Tina runs a hot dog stand. On Wednesday morning, she had 80 dozen hot dogs at her stand. The graph shows the number of hot dogs

KATRIN_1 [288]

Sunday: 45
Monday: 20
Tuesday: 0
All together: 65

6 0

2 years ago

Read 2 more answers

Describe the transformations of the following function. y=tan(2x-π)

Romashka [77]

Answer:

To quickly solve this problem, we can use a graphing tool or a calculator to plot the equation.

Please see the attached image below, to find more information about the graph

The equation is:

y = tan(2x - π)

y = tan (2*(x-π/2))

We can compare it to its parent function

g(x) = tan(x)

The answer is

Option a.

Horizontal shrink by 1/2 and horizontal shift of π/2 to the right

7 0

3 years ago

So every week in my class we do a riddle and if you’re able to solve 3 of these correctly in a row you get a prize (homework pas

BARSIC [14]

Answer:

18 but look at others awnsers aswell

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

The null hypothesis is accepted .

Assume the population variances are approximately the same

Step-by-step explanation:

Explanation:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

The first sample size 'n₁'= 20

mean of the first sample 'x₁⁻'= 17.53 pounds

standard deviation of first sample S₁ = 3.2 pounds

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

The second sample size n₂ = 24

mean of the second sample "x₂⁻"= 14.89 pounds

standard deviation of second sample S₂ = 2.7 pounds

Null hypothesis:-H₀: The Population Variance are approximately same

Alternatively hypothesis: H₁:The Population Variance are approximately same

Level of significance ∝ =0.05

Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

Test statistic :-

$t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

where $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}$

$S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}$

substitute values and we get S² = 40.988

$t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}$

t = 1.3622

Calculated value t = 1.3622

Tabulated value 't' = 2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

Conclusion:-

The null hypothesis is accepted

Assume the population variances are approximately the same.

6 0

2 years ago

PLEASE HELP MEEE!! THIS IS DUE IN THE NEXT 45 MINS PLEASEEE HELLPP :((

nikklg [1K]

They are all true None are false

3 0

3 years ago