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VladimirAG [237]
2 years ago
11

Find the equation of the straight line parallel to 2y=3x-7 and passing though the point (0.5,-1)

Mathematics
1 answer:
scZoUnD [109]2 years ago
8 0
<span>2y=3x-7 
y = 3/2x - 7/2
y = 1.5x - 3.5 so slope = 1.5
</span><span>line parallel so it has same slope = 1.5
</span>
y +1 = 1.5(x - 0.5)
y + 1 = 1.5x - 0.75
y = 1.5x  - 1.75

answer: equation y = 1.5x  - 1.75
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Assured normally costs £42 if you receive a 15% discount how much change will you get from £50
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5% of £42 = £2.10
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4. East High School has 540 students. There are 220 girls in the school.
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Answer:

11:16

Step-by-step explanation:

The ratio is 220:320.

Cancel the 0s then divide by 2.

22/2=11

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Can someone please help me with this? I don’t understand what is given.
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3 years ago
A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If
musickatia [10]

Answer:

a. The test statistic for this hypothesis would be z=1.71

b. Critical value at α = 0.10: z=1.282

c. Reject H0, the television network should not keep its current lineup.

d. H0: p ≤ 0.50; HA: p > 0.50

Step-by-step explanation:

We have to performa an hypothesis test on a proportion.

The claim is that the proportion of viewers that prefer the new show is bigger than 50% (meaning that most viewers prefer the new show).

The null hypothesis will state that both shows have the same proportion.

Then, the null and alternative hypothesis are:

H_0: \pi\leq0.50\\\\H_a: \pi>0.5

The sample proportion is p=0.53

p=X/n=438/827=0.53

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.5*0.5}{827}}= 0.017

Then, the z-statistic can be calculated as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.53-0.50-0.5/827}{0.017}=\dfrac{ 0.029 }{0.017} =1.706

The critical value for a right tailed test at a significance level of 0.10 is zc=1.282. This value can be looked up in a standard normal distribution table.

As the statistic is bigger than the critical value, it lies in the rejection region. The null hypothesis is rejected. There is evidence to support the claim that the proportion of viewers which support the new show is larger than 0.50.

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3 years ago
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