Susan made 2 galons of punch to by divided in 10 persons...how much punch did each person recieve

Zina [86]

Here, Just apply Distributive property which tells, a(b + c) = ab + bc
So, -6(-7 - 3k) = 42 + 18k

In short, Your Answer would be 42 +18k

Hope this helps!

4 0

3 years ago

Wires manufactured for use in a computer system are specified to have resistances between 0.11 and 0.13 ohms. The actual measure

Marina CMI [18]

Answer:

a) $P(0.11$

And we can find this probability with this difference and with the normal standard table or excel:

$P(-1.11$

b) $P(0.11 < \bar X < 0.13)$

And we can use the z score defined by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using the limits we got:

$z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22$

$z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22$

And we want to find this probability:

$P(-2.22< Z$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the resitances of a population, and for this case we know the distribution for X is given by:

$X \sim N(0.12,0.009)$

Where $\mu=0.12$ and $\sigma=0.009$

We are interested on this probability

$P(0.11$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(0.11$

And we can find this probability with this difference and with the normal standard table or excel:

$P(-1.11$

Part b

We select a sample size of n =4. And since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want this probability:

$P(0.11 < \bar X < 0.13)$

And we can use the z score defined by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using the limits we got:

$z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22$

$z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22$

And we want to find this probability:

$P(-2.22< Z$

4 0

3 years ago

Read 2 more answers

Jesse earns $12.62 an hour, plus she earns overtime for hours exceeding 40 hours per week. What is her gross weekly pay for work

san4es73 [151]

40(12.62) + 7(12.62(1.5)) = 504.80 + 7(18.93) = 504.80 + 132.51 =
637.31 <===

3 0

3 years ago

Read 2 more answers

Lines a and b are parallel. What is the measure of angle s? Enter your answer in the box. s = ∘∘ Horizontal parallel lines a and

VashaNatasha [74]

s = 22 degrees

Assuming I'm interpreting your description accurately, the top right angle of line b will also be 158 degrees because it's a corresponding angle to the top right angle of line a. That means that the top left angle of line b will be a supplementary angle. And supplementary angles add up to 180. So 180 - 158 = 22.

8 0

3 years ago

Read 2 more answers

Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final c

Black_prince [1.1K]

Answer:

The test statistic value Z = 4.64 > 1.96 at 0.05 level of significance

Null hypothesis is rejected

Alternative hypothesis is accepted

Test the claim that the proportion of all children in the town who suffer not from asthma is 11%

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 167

Given data In a sample of 167 children selected randomly from one town, it is found that 37 of them suffer from asthma.

Sample proportion

$p^{-} = \frac{x}{n} = \frac{37}{167} = 0.2215$

Given Population proportion 'P' = 11% = 0.11

Step(ii) :-

Null Hypothesis : H₀ : p = 11

Alternative Hypothesis H₁: p≠11

Test statistic

$Z = \frac{p^{-} -P }{\sqrt{\frac{P Q}{n} } }$

$Z = \frac{0.2215 -0.11 }{\sqrt{\frac{0.11 X 0.89}{167} } }$

Z = 4.64

step(iii)

Critical value Z = 1.96

The calculated value Z = 4.64 > 1.96 at 0.05 level of significance

Null hypothesis is rejected

Alternative hypothesis is accepted

Conclusion:-

Test the claim that the proportion of all children in the town who suffer not from asthma is 11%

4 0

4 years ago