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2 years ago

Jason's new business will start making a profit when he sells more than 20 items. Which graph best represents this situation?

Mathematics

2 answers:

masya89 [10]2 years ago

7 0

Answer:

I- where is the graph

Step-by-step explanation:

Anastaziya [24]2 years ago

6 0

Answer:

It's A the first graph

Step-by-step explanation:

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Which expression represents 43 minus the sum of 16 and 11

gogolik [260]

Answer:

43-(16*11) = -133

Step-by-step explanation:

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2 years ago

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PLEASE HELP ME, ill give points! Im in the 7th grade

sdas [7]

Answer: y=-5 x=5

explanation: look at the number the x and y is on

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2 years ago

PLEAS HELP!!!!!

Veseljchak [2.6K]

Answer: 11.59 months

Step-by-step explanation:

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3 years ago

If y=3x+5 and z=x^2+2x-5, what is yz?

Hitman42 [59]

<h3>Answer: 3x^3 + 11x^2 - 5x - 25</h3>

Work Shown:

yz = y(x^2+2x-5)

yz = y(x^2) + y(2x) + y(-5)

yz = x^2( y ) + 2x( y ) - 5( y )

yz = x^2( 3x+5 ) + 2x( 3x+5 ) - 5( 3x+5 )

yz = x^2*3x + x^2*5 + 2x*3x + 2x*5 - 5*3x - 5*5

yz = 3x^3 + 5x^2 + 6x^2 + 10x - 15x - 25

yz = 3x^3 + 11x^2 - 5x - 25

6 0

2 years ago

Use the normal distribution to find a confidence interval for a difference in proportions p1-p2 given the relevant sample result

9966 [12]

Answer:

a) $\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

b) $ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

c) $(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$

And the 99% confidence interval would be given (-0.319;0.0185).

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_1$ represent the real population proportion 1

$\hat p_1=0.2$ represent the estimated proportion 1

$n_1=60$ is the sample size required 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.35$ represent the estimated proportion 2

$n_2=100$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_2)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

Part a

The best estimate is given by:

$\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

Part c

And replacing into the confidence interval formula we got:

$(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$

And the 99% confidence interval would be given (-0.319;0.0185).

4 0

3 years ago