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3 years ago

Bobby decided to watch all of the movies in his

Mathematics

1 answer:

disa [49]3 years ago

4 0

Answer: C. y = 10x + 15

Step-by-step explanation:

He has already watched 15 movies so that's plus 15

And now he wants to watch 10 movies per week, we don't know how many weeks so that's 'x' in this equation.

If he watches movies for x weeks that's 10 times x which is the 10x in the equation.

Hope that helped!

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Leno4ka [110]

Answer:

$3, 661.60

Step-by-step explanation:

1703.44 + 1958.16

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3 years ago

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denis-greek [22]

Answer:

Slope= 2 / 10 = 1 / 5 = 0.2

y-intercept= (0, -4/5) or (0, -0.8)

Step-by-step explanation:

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4 years ago

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Murrr4er [49]

Answer:B

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3 years ago

In Abraham's last basketball game of the last season the ratio of free throws he made to the free throws he attempted was 12 ove

natita [175]

Answer:

Option C. $\frac{4}{6}$

Step-by-step explanation:

The complete question in the attached figure

we know that

The given ratio is equal to

$\frac{12}{18}$

Simplify

Divide by 3 both numerator and denominator

$\frac{12/3}{18/3}$

$\frac{4}{6}$

therefore

4/6 is an equivalent ratio of 12/18

3 0

3 years ago

A student at a four-year college claims that average enrollment at four-year colleges is higher than at two-year colleges in the

Kitty [74]

Answer:

a) Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

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

The degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$ Replacing the info given we got:

$t=\frac{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921$

c) $p_v =P(t_{68}>0.0921)=0.463$

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

Step-by-step explanation:

Information given

$\bar X_{1}=5069$ represent the mean for two year colleges

$\bar X_{2}=5216$ represent the mean for four year college

$s_{1}=4773$ represent the sample standard deviation for 1

$s_{2}=8141$ represent the sample standard deviation for 2

$n_{1}=35$ sample size for the group 2

$n_{2}=35$ sample size for the group 2

$\alpha=0.05$ Significance level provided

Part a

Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

Part b

We want to test if the mean of enrollment for four year college is higher than for two year college, the system of hypothesis would be:

Null hypothesis: $\mu_{2}-\mu_{1} \leq 0$

Alternative hypothesis: $\mu_{2} - \mu_{1} > 0$

The statistic is given by:

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

The degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$ Replacing the info given we got:

$t=\frac{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921$

Part c

The p value would be given by:

$p_v =P(t_{68}>0.0921)=0.463$

6 0

4 years ago