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Levart [38]
3 years ago
5

What is 0.50 as a fraction

Mathematics
2 answers:
SOVA2 [1]3 years ago
5 0
1/2 because the 0 after the 5 can be gone since it represent nothing. So, 5 is in the tenth place. So you put 5 over 10 (5/10). Then you can reduce it to 1/2.
ziro4ka [17]3 years ago
3 0
The answer is 50/100 same as the decimal sort of
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How do i find the area for number 4?
ankoles [38]
<h3>Answer:  41 cm^2</h3>

Explanation:

Form a horizontal line as shown below (see attached image). This forms 2 separate smaller rectangles

  • The bottom rectangle has area of 4*8 = 32 cm^2
  • The top rectangle has area of 3*3 = 9 cm^2

The total area is therefore 32+9 = 41 cm^2

You can say "square cm" or "sq cm" in place of the "cm^2".

5 0
3 years ago
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A report on consumer financial literacy summarized data from a representative sample of 1,664 adult Americans. Based on data fro
daser333 [38]

Answer:

a. The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

b. Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Sample of 1,664 adult Americans, 939 people in the sample would have given themselves a grade of A or B in personal finance.

This means that n = 1664, \pi = \frac{939}{1664} = 0.5643

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 - 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.54

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 + 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.588

The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

(b) Is the confidence interval from part (a) consistent with the statement that a majority of adult Americans would give themselves a grade of A or B?

Yes, because the confidence interval is entirely above 0.5.

Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

8 0
2 years ago
The diagram below shows a sign that Annie designed. It is made of two circles that share the same center. Annie wants to use str
Pavel [41]

132 in

Step-by-step explanation:

  • Take the total radius = 18+3 = 21= R (say)
  • Simply use the formula for the circumference i.e. 2*pi*R
  • 2*22/7*21= 132
3 0
2 years ago
Help pls it’s easy I’m just slow lol
IgorLugansk [536]

Answer:

12.65

Step-by-step explanation:

4 0
3 years ago
Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose P(V) = 0.17
ArbitrLikvidat [17]

Answer: 0.82

Step-by-step explanation:

The probability of the computer not containing neither a virus nor a worm is expressed as P(V^{C}∩W^{C}) , where P(V^{C}) is the probability that the event V doesn't happen and P(W^{C}) is the probability that the event W doesn't happen.

P(V^{C})= 1-P(V) = 1-0.17 = 0.83

P(W^{C})=1-P(W) = 1-0.05 = 0.95

Since V^{C} and W^{C} aren't mutually exclusive events, then:

P(V^{C}∪W^{C}) = P(V^{C}) + P(W^{C}) - P(V^{C}∩W^{C})

Isolating the probability that interests us:

P(V^{C}∩W^{C})= P(V^{C}) + P(W^{C}) - P(V^{C}∪W^{C})

Where P(V^{C}∪W^{C}) = 1 - 0.04 = 0.96

Finally:

P(V^{C}∩W^{C}) = 0.83 + 0.95 - 0.96 = 0.82

5 0
3 years ago
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