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3 years ago

James has a bag 77 of nickels and dimes. The total value of coins is $6.60. How many dimes does James have?

Mathematics

1 answer:

tatuchka [14]3 years ago

6 0

James has 55 dimes in the bag

Step-by-step explanation:

The given is:

James has in a bag 77 of nickels and dimes
The total value of coins is $6.60

We need to find how many dimes James has

Assume that the number of nickels is x and the number of

dimes is y in the bag

∵ There are x nickels in the bag

∵ There are y dimes in the bag

∵ There are 77 coins in the bag

∴ x + y = 77 ⇒ (1)

∵ 1 nickel = 5 cents

∴ The value of nickles in the bag = 5x cents

∵ 1 dime = 10 cents

∴ The value of dimes in the bag = 10y cents

∵ The total value of coins is $6.60

∵ 1 dollar = 100 cents

∴ $6.60 = 6.60 × 100 = 660 cents

∴ 5x + 10y = 660 ⇒ (2)

Now we have a system of equations to solve it

Multiply equation (1) by -10 to eliminate y

∵ -10x - 10y = -770 ⇒ (3)

- Add equations (2) and (3)

∴ -5x = -110

- Divide both sides by -5

∴ x = 22

Substitute the value of x in equation (2) to find y

∵ 22 + y = 77

- Subtract 22 from both sides

∴ y = 55

∴ The number of dimes is 55

James has 55 dimes in the bag

Learn more:

You can learn more about the system of equations in brainly.com/question/6075514

#LearnwithBrainly

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Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

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(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

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where, $\bar X$ = sample mean mg of mercury = 3.59

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So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

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Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

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