Which equation, in point-slope form, passes through (-2, 4) and has a slope of 3?

Alecsey [184]

I am pretty sure it is 30. Hope it helps!!

3 0

3 years ago

Mark orders food for a restaurant. He orders 5 times the number of pounds of chicken as he does beef, and he orders 4 times the

tester [92]

180 pounds of chicken and 225 of beef

8 0

3 years ago

Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak

crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0

2 years ago

A boat traveled 128 kilometers in

AlekseyPX

Answer:

The Distance cover by boat in 2 hours and 15 minutes is 72 km

Step-by-step explanation:

Given as :

The Distance cover by boat = 128 kilometer

The time taken is 4 hours

As , Distance = Speed ×Time

So , Speed = $\frac{Diastance}{Time}$

Or, Speed = $\frac{128 km}{4 hours}$

∴ Speed = 32 km per hour

Now, again The boat travel 2 hours and 15 minutes further with speed 32 kmph

So, Time2 = 2 + $\frac{15}{60}$

Or, Time2 = 2 + $\frac{1}{4}$

I.e Time2 = $\frac{9}{4}$ hours

Again , Distance = Speed ×Time

So , Distance = 32kmph × $\frac{9}{4}$ hours

I.e Distance = $(\frac{32\times 9}{4})$ = 72 kilometer

Hence The Distance cover by boat in 2 hours and 15 minutes is 72 km Answer

5 0

2 years ago

Hozumi recorded the daily high temperatures for an 18-day period in the table shown please I need this really badly

mr Goodwill [35]

Answer:

the answer is letter A

Step-by-step explanation:

7 0

3 years ago