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tamaranim1 [39]
3 years ago
15

You know maths yesterday homework​

Mathematics
1 answer:
dalvyx [7]3 years ago
5 0

Answer:

Explain what you mean please :)?

Step-by-step explanation:

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irga5000 [103]

Answer:

y=1.3 x - 0.21

Step-by-step explanation:

y = (slope) \times x + (constant)

set (-2.3,-3.2) in above

- 3.2 = 1.3 \times ( - 2.3) + c

then c=-0.21

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2 years ago
What is the value of the expression below when y=8 2y+7
Gala2k [10]

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23

Step-by-step explanation:

if y=8 then you plug it inti the equation

2(8)+7

16+7=23

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Anon25 [30]

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7.375 ib.

Step-by-step explanation:

You just add it all together

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2 years ago
Use the information to answer the question.
Alona [7]

Answer:

13x^8 + 4x^5 - 16x^2 try this

Step-by-step explanation:

8 0
3 years ago
A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he
UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

For this problem, we have that:

We dont know the true proportion, so we use \pi = 0.5, which is when we are are going to need the largest sample size.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.04}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2

n = 600.25

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.02\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.02}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2

n = 2401

The minimum sample size is 2401.

4 0
3 years ago
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