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3 years ago

Can someone help me please ?

Mathematics

1 answer:

STatiana [176]3 years ago

4 0

Answer:

Answer choice D, 8 1/2

Step-by-step explanation:

Covert 3 3/16 to an improper fraction. It would be 51/16. Now if we divide witha fraction, we use the reciprocal of the fraction. 51/16 * 8/3 = 408/48. This is also equal to 8.5 or 17/2. So the correct answer would be 8 1/2.

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You are playing a game where you draw a card from a standard deck, and you will win $13 if you draw a face card,

marissa [1.9K]

Answer:

Step-by-step explanation:

win

There are 3 face cards / suit. Every suit contains a Jack, Queen, and king.

There are 4 suits so the number of face cards is 3*4 = 12

There is 1 ace / suit

There are 4 suits so the number of aces is 1 * 4 = 4

The gain from the face cards = 12 * 13 = 156

The gain from the aces = 4 * 3 = 12

Total gain = 168

Losses

Any other card costs you $5

How many cards is that? 52 - 16 = 36

Each one costs 5 dollars

Total loss = 5 * 36 = 180

So each time you play the game, you can expect to lose 180 - 68 = 12 dollars. You should walk away quickly from this one.

8 0

3 years ago

The mayor of a town has proposed a plan for the construction of an adjoining bridge. A political study took a sample of 13001300

Ket [755]

Answer:

No, there is not enough evidence at the 0.02 level to support the strategist's claim.

Step-by-step explanation:

We are given that a political study took a sample of 1300 voters in the town and found that 57% of the residents favored construction.

And, a political strategist wants to test the claim that the percentage of residents who favor construction is more than 54%, i.e;

Null Hypothesis, $H_0$ : p = 0.54 {means that the percentage of residents who favor construction is 54%}

Alternate Hypothesis, $H_1$ : p > 0.54 {means that the percentage of residents who favor construction is more than 54%}

The test statistics we will use here is;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, p = actual percentage of residents who favor construction = 0.54

$\hat p$ = percentage of residents who favor construction in a sample of

1300 voters = 0.57

n = sample of voters = 1300

So, Test statistics = $\frac{0.57 -0.54}{\sqrt{\frac{0.57(1- 0.57)}{1300} } }$

= 2.185

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the percentage of residents who favor construction is 54% and the strategist's claim is not supported.

4 0

4 years ago

2x = 4y + 8, x - 2y = 4

Evgen [1.6K]

Answer: x = 8, y = 2, coordinates (8,2)

Step-by-step explanation:

2x = 4y + 8 (divide by 2 to get value of x)

x = 2y + 4

(now substitute x into the second equation to find value of y)

2y + 4 - 2y = 4

2y - 2y = 4 - 4 (divide by 2 to get value of y)

y - y = 2 - 2

y = 2

Now plug in value of "y" to get value of "x".

x - 2(2) = 4

x - 4 = 4

x = 8

Coordinates (8,2)

3 0

3 years ago

Maths substitution y=2x-10<br> Y=-4x+8

joja [24]

8 0

3 years ago

Read 2 more answers

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