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kap26 [50]
3 years ago
10

Tarana mailed 20 greetings cards. 9 of them were mailed to USA. What percentage of the cards were mailed to USA?

Mathematics
1 answer:
Elza [17]3 years ago
7 0

Answer:

45%

Step-by-step explanation:

9/20=0.45 (Divide it first), 45/100=45% (45/100 because 100 makes 1 whole -all 20 cards-)

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The diagonal should be 3 feet
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Please show work and correct answers ONLY!
amm1812

2.

April 11 : 3 hours 15 minutes

April 12 : 3 hours

April 13 : 4 hours 30 minutes

April 14 : 5 hours 15 minutes

April 15 : 2 hours 45 minutes

total : 18 hours 45 minutes

10.50 * 18.75 = $196.875

answer: $196.88

3.

45 1/4-> 45.25 * .5 = 22.625 = 22 5/8

gusset plate :

5 3/4 -> 5.75 * .5 = 2.875 = 2 7/8

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2 years ago
Number 17 thanks pls ASAP ASAP
RideAnS [48]

Answer:

\frac{1}{2}  which agrees with answer B

Step-by-step explanation:

First write the equation that represents this type of variation:

y=\frac{5}{x}

then we need to solve for "x" when y  = 10 as shown below:

y=\frac{5}{x} \\10=\frac{5}{x} \\10*x=5\\x=\frac{5}{10} \\x=\frac{1}{2}

5 0
3 years ago
1. X - 18 = 23
Thepotemich [5.8K]

Answer:

  1. 41
  2. -5
  3. 42
  4. 21
  5. 4
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Step-by-step explanation:

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3 years ago
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An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos
Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

8 0
3 years ago
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