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3 years ago

100 POINTS PLEASE HELP ASAP!!!!!

Mathematics

1 answer:

UkoKoshka [18]3 years ago

7 0

The two numbers are 31 and 35.

Step-by-step explanation:

Let,

one number = x

2nd number = y

According to statement;

x+y=66 Eqn 1

y=x+4 Eqn 2

Putting y=x+4 from Eqn 2 in Eqn 1;

$x+(x+4)=66\\x+x+4=66\\2x=66-4\\2x=62\\Dividing\ both\ sides\ by\ 2\\\frac{2x}{2}=\frac{62}{2}\\x=31\\Putting\ x=31\ in\ Eqn\ 2\\y=31+4\\y=35$

The two numbers are 31 and 35.

Keywords: Linear equation, Addition

Learn more about linear equations at:

#LearnwithBrainly

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Semmy [17]

Answer: 6

explanation:
set up proportions
(4+2)/9 = 4/x
6/9 = 4/x
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2/3 = 4/x
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8 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

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3 years ago

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