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3 years ago

Please Help! I'll give Brainliest to whoever gets the answer

Mathematics

2 answers:

Svetradugi [14.3K]3 years ago

4 0

Answer:

shrimp cocktail: 0.00 or 0%

lifesaver: 0.25 or 25%

gum: 0.05 or 5%

chocolates or lifesaver: 0.80 or 80%

Not Cracker: 0.85 or 85%

Any treat: 1.00 or 100%

Step-by-step explanation:

First Add all of them together

22+10+6+2=40

now divide

22/40 0.55

10/40=0.25

6/40=0.15

2/40=0.05

add/subtract

0.25+0.55=80%

1-0.15=0.85

Hope this helps! Plz award Brainliest : )

Harman [31]3 years ago

3 0

Answer:

The other guy is right

Step-by-step explanation:

give the man Brainliest

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How would I solve the equations: y = x-4 and y=x^2-4x using substitution

andriy [413]

Answer:

x = 6

Explanation:

y = x - 4

26 = 4x + y

If y = y, then we can rewrite this system as this: 4x + x - 4 = 26

Then we combine like terms and add 4 on both sides to give us 5 x = 30 . Divide by 5 on both sides and we arrive at the solution: x = 6

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3 years ago

Help me please, you get 50 points. This is easy I just need help

pentagon [3]

Fraction -> Decimal

4/5 = 0.8

1/8 = 0.125

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0.22 * 100% = 22%

0.47 * 100% = 47%

Percent -> Fraction

34% / 100 = 0.34 = 17/50

25% / 100 = 0.25 = 1/4

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1/3 = 0.33... * 100% = 33.3%

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Decimal -> Fraction

0.65 = 65/100 = 13/20

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3 years ago

5. Five-eighth of a number is 15. What is the number?<br> Your answer

anygoal [31]

Answer:

I think it is 24

Step-by-step explanation:

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3 years ago

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Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

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3 years ago

What value for n makes the equation true? 8n - 2 = 2

kiruha [24]

Answer:

n = 1/2

Step-by-step explanation:

The answer is 1/2

I hope this helps you

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