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faust18 [17]
3 years ago
5

harold was 5 years older than Mary. In 10 years he found that 4 times his age exceeded twice Mary's age by only 50. How old were

Harold and Mary in the beginning?
Mathematics
1 answer:
Ivanshal [37]3 years ago
8 0

in the beginning 

Harold X+5
Mary X

In 10 years

Harold X+5+10 X+15
Mary X+10 

4(X+15)-2(X+10)=50
4X+60-2X-20=50
2X+40=50
2X=50-40
2X=10
X= 10÷2
X=5

Harold At The Beginning 5+5=10
Mary At The Beginning 5
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In a standard deck of cards, what is the probability of drawing a red card or a face card? answer choices are in the form of a p
Orlov [11]

In a standard set of cards, the probability of drawing a red card or a face card is  73%

<h3>What is the probability?</h3>

In a standard deck of cards, there are 26 red cards. In that same deck, there will be 12 face cards.

The probability of picking either one is:

= (Number of face cards + Number of red cards) / Number of total cards

Solving gives:

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4 0
2 years ago
PLEASE HELP!!!
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c²(1 + 3d)

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c² + d

Twice the product of c and d is;

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c² + d + 2c²d

or

c²(1 + 3d)

8 0
3 years ago
The strength of a certain type of rubber is tested by subjecting pieces of the rubber to an abrasion test. For the rubber to be
My name is Ann [436]

Answer:

Step-by-step explanation:

Hello!

You have the hypothesis that the average weight loss for rubber after an abrasion test is less than 3.5 mg. To test this a large sample of pieces of rubber were sampled and subjected to the abrasion test.

With the given information you must test whether the researcher's hypothesis is sustained or not.

The study variable is,

X: Weight loss of rubber cured in a certain way after being subjected to the abrasion test. (mg)

There is no information about the variable distribution, but since it is said that the sample is a "large number" I'll take it as if it is bigger than 30 and apply the Central Limit Theorem to use the approximation of the sample mean to normal. This way I can use the Z-statistic for the test.

Symbolically the statistic hypothesis is:

H₀: μ ≥ 3.5

H₁: μ < 3.5

α: 0.05 (since is not listed, I'll choose one of the most common signification levels)

You have a one-tailed critical region, this means the p-value will also be one-tailed to the left of the distribution (i.e. →-∞)

The formula of the statistic is:

Z= <u> X[bar] - μ </u> ≈ N(0;1)

        δ/√n

To calculate the statistic you have to use the information given.

The sample mean X[bar]= 3.4 mg

Upper bond of 95% CI= 3.45 mg

The basic structure of a CI for the mean is

"estimator" ± "margin of error"

Upper bound is "estimator" + "margin of error"

Using the formula:

Ub= X[bar] + d ⇒ 3.45= 3.4 + d

⇒ d= 3.45 - 3.4 = 0.05

Where d is the margin of error

d= Z_{1-\alpha /2} * (δ/√n)

d= Z_{0.975} * (δ/√n)

d/Z_{0.975}= (δ/√n)

(δ/√n)= 0.05/ 1.96 = 0.0255

(δ/√n) is the denominator in the formula, corresponds to the standard deviation of the distribution.

Now you have all values and can calculate the statistic under the null hypothesis:

Z= <u> 3.4 - 3.5 </u> = -3.92

       0.0255

And the p-value:

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To calculate the exact probability I've used a statistic program.

p-value < 0.001

I hope it helps!

8 0
3 years ago
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