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lara31 [8.8K]
2 years ago
10

Between which pair of numbers is the exact product of 379 and 8?

Mathematics
2 answers:
klasskru [66]2 years ago
4 0

The complete question is

Between which pair of numbers is the exact product of 379 and 8?

A between 2,400 and 2,500

B between 2,400 and 2,800

C between 2,400 and 3,000

D between 2,400 and 3,200

Find the product

Multiply 379 by 8

379*8=3,032

therefore

2,400

the answer is the option

D between 2,400 and 3,200

nata0808 [166]2 years ago
3 0
<span>379 times 8 is 3032. This result, 3032 is between 3031 and 3033, so the answer is that the exact product of 379 and 8 is between 3031 and 3033.</span>
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The total tax rate for each plan is computed below:
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Plan A= $9450
For Plan B, tax is computed at 10% across all earnings
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Plan B= $9800
Difference = Plan B-Plan A=($9800-$9450)
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From the computed tax rate in each plan. Plan B will pay $350 more than Plan A.

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What value of n makes the statement true? 6xn · 4x2 = 24x6
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We check first the numerical coefficients of both sides of the equation if they match if we perform the operation.
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Then, the variables. For multiplication with the same variables, the exponents are added. In the given above,
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Which statements are true about the location of 3.28 on the number line? Check all that apply. A number line from 1.2 to 4.2. 3.
Juliette [100K]

Answer:

- 3.28 should be plotted between 3.2 and 3.4

- 3.28 is closer to 3.0 than 4.0.

- 3.28 is closer to 3.2 than 3.4.

Step-by-step explanation:

3.28 is located towards the positive side of the number line being a positive value. Since the value is located between 3.2 and 3.4, therefore it can be plotted between this two points.

Also 3.28 is known to be closer to 3.0 than 4.0 because the difference between 3.28 and 3.0 is lower than the difference between 3.28 and 4.

4-3.28 = 0.72(larger value)

3.28-3.0 = 0.28 (smaller value)

The smaller the difference, the closer the value of 3.28 to the value in consideration.

Similarly, 3.28 is closer to 3.2 than 3.4, due to their differences. The difference between 3.28 and 3.2 is lower than the difference between 3.28 and 3.4 as shown:

3.28 - 3.2 = 0.08(smaller)

3.4-3.28 = 0.12(larger)

8 0
2 years ago
Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

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5 0
2 years ago
Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4%
bogdanovich [222]

Mr. Jackson invested $800 at 6% per year and $ 2400  at 4 % per year

<h3><u>Solution:</u></h3>

Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4% per year.

Let the sum invested be ‘a’ and ‘3a’ at 6% per year and 4 % per year respectively

Also, his annual return totaled $144

We can form following equation on the basis of question:-

\begin{array}{l}{\text { Then, } \frac{a \times 6 \times 1}{100}+\frac{3 a \times 4 \times 1}{100}=\$ 144} \\\\ {\frac{6 a}{100}+\frac{12 a}{100}=144} \\\\ {\frac{6 a+12 a}{100}=144} \\\\ {\frac{18 a}{100}=144} \\\\ {18 a=14400} \\\\ {a=14400 \div 18}\end{array}

a = $800

The amount of money invested at 6% = a = 800

The amount of money invested at 4 % = 3a = 3(800) = 2400

So, the amount of money invested at 6% is $800 and the amount of money invested at 4% is $ 2400

4 0
2 years ago
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