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inessss [21]
4 years ago
10

Can any one tell me if this is correct or is it translation

Mathematics
2 answers:
Anastasy [175]4 years ago
7 0
No, it’s not a reflection. It’s a translation.
In-s [12.5K]4 years ago
4 0
Yes I believe it is teflection
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Work out, giving your answer in its simplest form:<br> 4/7 divided by 2 1 / 3
Vitek1552 [10]

Step-by-step explanation:

4/7÷21/3

=4/7*3/21

=12/147

8 0
3 years ago
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A milkshake recipe calls for 1/3 cup of milk for each milkshake. If a server has 15/8 cups of milk, what is the maximum number o
Zepler [3.9K]

Answer:

5 milk shakes

Step-by-step explanation:

From the above question, we know that for the recipe

1/3 cup of milk = 1 milk shake

15/8 cup of milk = y milk shakes

We cross Multiply

15/8 cup × 1 = 1/3cup × y

y = 15/8 cup × 1 /1/3cup

= 15/8 ÷ 1/3

= 15/8 × 3

= 45/8

= 5.625 of 5 5/8 cups of milk

Therefore, the maximum number of

milkshakes the server can make following the recipe porpotians is 5 milk shakes.

4 0
4 years ago
Which of the following is the point-slope form of the line?
ryzh [129]

Answer:

A

Step-by-step explanation:

We can see that the slope is positive, which means that the x-term must be positive.

If you expand and simply both equations into the form y = mx + b, you will find that m is positive for A and negative for B, hence A is the correct answer.

4 0
3 years ago
A company manufactures televisions. The average weight of the televisions is 5 pounds with a standard deviation of 0.1 pound. As
Semenov [28]

Answer:

0.2564\text{ pounds}

Step-by-step explanation:

The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.

To find the X percentile for the television weights, use the formula:

X=\mu +k\sigma, where \mu is the average of the set, k is some constant relevant to the percentile you're finding, and \sigma is one standard deviation.

As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute \mu=5, k=1.282, and \sigma=0.1:

X=5+(1.282)(0.1)=5.1282

Therefore, the 90th percentile weight is 5.1282 pounds.

Repeat the process for calculating the 10th percentile weight:

X=5+(-1.282)(0.1)=4.8718

The difference between these two weights is 5.1282-4.8718=\boxed{0.2564\text{ pounds}}.

8 0
3 years ago
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A pound of crackers costs $2.88. You purchased 1/4 pound of crackers. How much money did you pay?
Maru [420]

Answer:

72 cents

Step-by-step explanation:

8 0
3 years ago
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