All categories

2 years ago

Comedy Drama Action Horror

Mathematics

1 answer:

Neporo4naja [7]2 years ago

4 0

Answer:

B. Drama is the most popular genre

Step-by-step explanation:

⇒ Comedy: 5

Drama: 7

Action: 3

Horror: 1

⇒ Comedy: 2

Drama: 11

Action: 4

Horror: 5

<h3 /><h3>With this information we can start ruling out our answer choices.</h3>

<u>A. horror is the most popular genre</u>

Lets see, both of the sum of both days for <u>Horror is 6</u>, for <u>Action is 7,</u> <u>Drama is 18,</u> and <u>Comedy is 7, </u>with this we can conclude that<u> </u><u><em>horror is actually the least popular genre</em></u>

<u>B. Drama is the most popular genre</u>

Using the information we gathered previously, we know <u>Drama is the most popular genre</u>, therefore this option is a high possibility.

<u>C. People like comedies because they like to laugh</u>

Although, I personally love comedies for this reason, <u>this chart does not prove that people like comedies for this reason.</u>

<u>D. People don't like horror movies because they don't like to be scared</u>

As for option C, the <u>chart does not prove that is the case!</u>

<h3>Therefore we can conclude that <u>option B</u> is the best choice.</h3>

You might be interested in

0.1562 rounded to the nearest hundredth place

kondaur [170]

Answer:

0.16

Step-by-step explanation:

if the number in the thousandths is 5 or over you round it up.

3 0

3 years ago

Read 2 more answers

How Many Solutions?<br> – 3 (v + 4) = 2v - 37<br> Infmute Solution<br> No Solution<br> One Solution

Olegator [25]

There is only one solution

V=5

6 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

For what value of h is the expression (x-h)^2+6 equivalent to the expression x^2-6x+15

Yuri [45]

Answer:

1. h=3 2. h=1

Step-by-step explanation:

You need to make both sides into perfect square trinomials that you factor into the perfect square of a binomial. The left side is almost there to start with.

ax² + bx + c is the standard equation

1. ( x – h)² + 6 = x² - 6x + 15 subtract 6 from both sides to get ( x – h)² alone

- 6 - 6

(x – h)² = x² - 6x + 9 factor the right side into a perfect square (b/2)

here b = -6 so b/2 = -3 so (x – 3)²

(x – 3)² = (x - 3)(x - 3) = FOIL(first, outside, inside, last)

(x)(x) + (-3)(x) + (-3)(x) + (-3)(-3) = x² - 3x - 3x + 9

= x² - 6x + 9 = (x – 3)² so

(x – h)² = (x - 3)² so h= 3

ax² + bx + c is the standard equation

2. ( x – h)² - 6 = x² - 2x - 5 add 6 to both sides to get ( x – h)² alone

+ 6 + 6

(x – h)² = x² - 2x + 1 factor the right side into a perfect square (b/2)

here b = -2 so b/2 = -1 so (x – 1)²

(x – 1)² = (x - 1)(x - 1) = FOIL(first, outside, inside, last)

(x)(x) + (-1)(x) + (-1)(x) + (-1)(-1) = x² - 1x - 1x + 1

= x² - 2x + 1 = (x – 1)² so

(x – h)² = (x - 1)² so h= 1

4 0

3 years ago

Which set of data has the smallest difference between the mean and the median?

Andre45 [30]

Answer:

D 23, 36, 25, 38, 30

Step-by-step explanation:

Mean of a data-set:

The mean of a data-set is the sum of all it's values divided by the number of values.

Median of a data-set:

The median of a data-set is the value that separates the lower 50% from the upper 50% on the ordered set.

If the cardinality n is odd, as in this case, the median is the element at the position (n+1)/2. Since in this case, n = 5, the median is the 3rd element in the ordered set.

A 13, 42, 17, 26, 38

Ordered set: 13, 17, 26, 38, 42

Median: 26

Mean: (13+17+26+38+42)/5 = 27.2

Difference: 27.2 - 26 = 1.2

B. 28, 7, 36, 8, 17

Ordered set: 7, 8, 17, 28, 36

Median: 17

Mean: (7+8+17+28+36)/5 = 19.2

Difference: 19.2 - 17 = 2.2

C 14, 29, 38, 7, 13

Ordered set: 7, 13, 14, 29, 38

Median: 14

Mean: (7+13+14+29+38)/5 = 20.2

Difference: 20.2 - 14 = 6.2

D 23, 36, 25, 38, 30

Ordered set: 23, 25, 30, 36, 38

Median: 30

Mean: (23+25+30+36+38)/5 = 30.4

Difference: 30.4 - 30 = 0.4

Option d has the smallest difference.

6 0

3 years ago