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3 years ago

6

A movie theater charges $10.50 per ticket and $5 per bucket of popcorn. You never buy the popcorn because you think it's too exp

ensive.
Is the total price you pay proportional to the number of tickets you buy?

2 answers:

KengaRu [80]3 years ago

5 0

Yes I think so I could be wrong but I think the answer is yes it is proportional

White raven [17]3 years ago

5 0

Yes It is proportional

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I do have examples to help me but they make no sense :/

RideAnS [48]

First you would do 5^2 and that is 25. Next do 75+25 and that is 100. So I just did the top part next I would do the bottom. We know 5^2 is 25 and know we just divide 100 by 25 and that is 4. So your answer is 4

7 0

2 years ago

Time sensitive mark best as brainiest

ycow [4]

Answer:

4986 ounces/m³

Step-by-step explanation:

1 kilogram = 35.274 ounces

1 cubic foot = 0.0283 cubic metre

We are converting kg/ft³ to ounces/m³

Hence:

4kg/ft³ × 35.274 ounces/ 1 kg × 1 ft³/0.0283m³

= 4985.7243816 ounces/m³

Approximately to the nearest whole number = 4986 ounces/m³

6 0

3 years ago

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0

3 years ago

The graphs below show measurements from cubes

UNO [17]

Answer:

the first one and the third one

Step-by-step explanation:

5 0

3 years ago

Decrease £1200 by 25%

gayaneshka [121]

Ok well...
1200 times .25 is 300 so subtract 1200-300= 900 is the decreased number :)

Hope it helps! :)

6 0

4 years ago