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Marina86 [1]
3 years ago
14

PLEASE HELP IM BEGGING YOU!!! Team Integrity kids were supposed to go on a field trip to an ice rink. The cost was $12 per stude

nt for tickets, and $10 per student for food. The team paid $50 for buses. Write an expression to represent the total cost of this field trip for the entire team. Identify what your variable represents. *
Mathematics
1 answer:
IceJOKER [234]3 years ago
5 0
(12+10)s+50= t
S=students
T= total
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Determine whether the relationship between the circumfrance of a circle and its diameter is a direct variation. If so, identify
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Answer:

The relationship between the circumference of a circle and its diameter represent  a direct variation and the constant of proportionality is equal to the constant \pi

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form y=kx

where K is the constant of proportionality

In this problem we know that

The circumference of a circle is equal to

C=\pi D

therefore

the relationship between the circumference of a circle and its diameter is a direct variation and the constant of proportionality is equal to the constant \pi

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3 years ago
What is 2 5/6 + 7/4 <br> A. 2 2/5<br> B. 3 1/5<br> C. 3 4/5<br> D. 4 7/12
Mazyrski [523]

Answer:

D

Step-by-step explanation:

2 5/6=2 10/12

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2 10/12+1 9/12= 4 7/12

you can just change the denominators to the same which is 12. Hope you appreciate it!

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3 years ago
Plz help you have to find the angle marked x in each of these polygons
svetoff [14.1K]
Https://us-static.z-dn.net/files/dd0/5d5ddccb833d70732c67b84ec119f42f.jpeg

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Answer:

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4 0
3 years ago
According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating
Natali [406]

Answer:

a) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

H_0: \mu=32.48\\\\H_a:\mu< 32.48

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565

The degrees of freedom for this sample size are:

df=n-1=65-1=64

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

<u>Critical value approach</u>

<u></u>

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

6 0
3 years ago
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