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Marina CMI [18]
2 years ago
5

Colton has 8 comedy movies, 6 action movies, 555 fantasy movies, and 1 drama movie on his list of favorite movies.

Mathematics
2 answers:
11Alexandr11 [23.1K]2 years ago
8 0

Answer:

The correct option is A.

Step-by-step explanation:

It is given that Colton has 8 comedy movies, 6 action movies, 5 fantasy movies, and 1 drama movie on his list of favorite movies.

Total number of favorite movies is

8+6+5+1=20

Therefore Colton has 20 movies in the list of favorite movies.

For every 5 movies on the list, there are 2 movies of required category.

\frac{2}{5}\times \frac{4}{4}=\frac{8}{20}

For every 20 movies on the list, there are 8 movies of required category.

Out of 20 movies we have 8 comedy movies, therefore the required category is comedy movie.

The ratio of comedy movie and total movies is

\frac{8}{20}=\frac{2}{5}

Therefore we can say that for every 5 movies on the list, there are 2 comedy movies. Option A is correct.

zhuklara [117]2 years ago
3 0

Answer:

A

Step-by-step explanation:

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Vera_Pavlovna [14]

Answer:

x = 14      D

Step-by-step explanation:

(18x + 36) / 3 = 6x + 12

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8 0
2 years ago
Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor’s. A sample of 40 bab
Leviafan [203]

Answer:

z=\frac{(7.6-8.1)-0}{\sqrt{\frac{2.3^2}{40}+\frac{2.9^2}{55}}}}=-0.936  

The p value can be founded with this formula:

p_v =P(z  

Since the p value is higher than the significance level provided of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean for the Gibbs brand is significantly lower than the true mean for the competitor

Step-by-step explanation:

Information given

\bar X_{1}=7.6 represent the mean for Gibbs products

\bar X_{2}=8.1 represent the mean for the competitor

\sigma_{1}=2.3 represent the population standard deviation for Gibbs

\sigma_{2}=2.9 represent the sample standard deviation for the competitor

n_{1}=40 sample size for the group Gibbs

n_{2}=55 sample size for the group competitor

\alpha=0.05 Significance level provided

z would represent the statistic

Hypothesis to verify

We want to check if babies using the Gibbs brand gained less weight, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}=0  

Alternative hypothesis:\mu_{1} - \mu_{2}< 0  

The statistic would be given by:

z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

Replacing the info given we got:

z=\frac{(7.6-8.1)-0}{\sqrt{\frac{2.3^2}{40}+\frac{2.9^2}{55}}}}=-0.936  

The p value can be founded with this formula:

p_v =P(z  

Since the p value is higher than the significance level provided of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean for the Gibbs brand is significantly lower than the true mean for the competitor

5 0
3 years ago
is vector v with an initial point of (-5,22) and a terminal point of (20,60) equal to vector u with an intintal point of (50,120
marta [7]

Answer:

Yes, vectors u and v are equal.

Step-by-step explanation:

We need to check whether vectors u and v are equal or not.

If the initial point is (x_1,y_1) and terminal point is (x_2,y_2), then the vector is

Vector=(x_2-x_1)i+(y_2-y_1)j

Vector v with an initial point of (-5,22) and a terminal point of (20,60).

\overrightarrow v=(20-(-5))i+(60-22)j

\overrightarrow v=25i+38j       ..... (1)

Vector u with an initial point of (50,120) and a terminal point of (75,158).

\overrightarrow u=(75-50)i+(158-120)j

\overrightarrow u=25i+38j         .... (2)

From (1) and (2) we get

\overrightarrow u=\overrightarrow v

Therefore, vectors u and v are equal.

5 0
3 years ago
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