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natka813 [3]
3 years ago
5

Find 50% of$199.96.(Round to the nearest penny.)​

Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

$99.98

Step-by-step explanation:

199.96 x .5 = 99.98

.5=50% so if you multiply by .5 you get 50 percent of 99.98

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This one show a positive relationship between x axis and y axis, as the x goes up, the y goes up too.

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2 years ago
The length of a rectangle is 6 yd more than twice the width x. The area is 416 yd2. Find the dimensions of the rectangle.
kati45 [8]

Answer:

13 yds * 32 yds

Step-by-step explanation:

A=area, l=length, w=width

A=w*l

l=6+2w

A=w*(6+2w)

A=6w+2w^2

416=6w+2w^2

416/2=(6w+2w^2)/2

208=3w+w^2

208-208=3w+w^2-208

w^2+3w-208=0

w^2+16w-13w-208=0

w(w+16)-13(w+16)=0

(w-13)(w+16)=0 ==> w+16=0 ==> w=-16, w can't be negative.

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<u>w=13</u>

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l=6+2(13)

l=6+26

<u>l=32</u>

Dimensions: 13 * 32

8 0
9 months ago
Helppppp plsss<br> Ill mark brainliest 20pts
Citrus2011 [14]
A) For both sets A and B, calculating the mean, range, and quartiles are a good way of measuring the center and spread. Using standard deviation may not be the best because we do not know whether the distributions are normal or not.

b) For set A, the lowest value is 63, while the highest is 86. An estimate for the mean, based on the average of these, is 74.5. Most of the 70+ values are below 74.5, so we may guess that the mean will be above the median.
For set B, the lowest is 63, while the maximum is 95. The estimated mean would be 79. But since there are more values on the 80+ and 90+ side, the median is likely to be higher than 79.

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2 years ago
5.11.
irinina [24]

Answer:

Following are the response to the given points:

Step-by-step explanation:

For question 5.11:

For point a:

For all the particular circumstances, it was not an appropriate sampling strategy as each normal distribution acquired is at a minimum of 30(5) = 150 or 2.5 hours for a time. Its point is not absolutely fair if it exhibits any spike change for roughly 10 minutes.

For point b:

The problem would be that the process can transition to an in the state in less than half an hour and return to in the state. Thus, each subgroup is a biased selection of the whole element created over the last 2 \frac{1}{2} hours. Another sampling approach is a group.

For question 5.12:

This production method creates 500 pieces each day. A sampling section is selected every half an hour, and the average of five dimensions can be seen in a \bar{x}line graph when 5 parts were achieved.

This is not an appropriate sampling method if the assigned reason leads to a sluggish, prolonged uplift. The difficulty would be that gradual or longer upward drift in the procedure takes or less half an hour then returns to a controlled state. Suppose that a shift of both the detectable size will last hours 2 \frac{1}{2} . An alternative type of analysis should be a random sample of five consecutive pieces created every 2 \frac{1}{2} hour.

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2 years ago
Use the X method to find the solutions of<br><br> 6x2 + 2x – 20 = 0
SVEN [57.7K]
The X method?
2x=8
X=4
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3 years ago
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