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bonufazy [111]
4 years ago
7

What is 149 times 5. solve and estimate

Mathematics
2 answers:
kiruha [24]4 years ago
8 0
149 times 5
9 times 5 equals 45 carry the 4 4x5 equals 20 add the 4 you carried 5 times 1 is 5 Plus 2 equals 7 which is 745 745 is your main answer and you could either round this up or down since it's right in the middle so 740 or 7:50 would work so does your estimate so basically that your answer
GenaCL600 [577]4 years ago
6 0
The answer to this problem is 745.

149 * 5 = 745

In the future, you can do this with a calculator.
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VMariaS [17]

Answer:

Step-by-step explanation:

i hope i am right q=16 is my answer my bad if it is wrong

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Please answer the question above.
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3 years ago
Solve for x . x/2 + 1.1= -2.1
raketka [301]

Answer: x = 1.6

Step-by-step explanation:

Start by Identifying what you need to keep, in this case that would be X. In order to solve for x, you would need to get rid of any co-efficients. (example: +1.1) To get rid of them, you have to do the opposite action of what it's doing to the equation. (subtraction)

1.1- (-2.1) = 3.2 (if you subtract a negative, it turns positive/ you add) leaving you with x/2 = 3.2.

Now that we have our coefficient out the way, we have to get rid of the division so we can <u>only</u> solve for x. We do that by dividing the remainder of the equation by 2

3.2/2 = 1.6

Leaving us with the answer, x = 1.6

8 0
3 years ago
What is the difference 1/8 - 3/10 a.17/40 b.7/40 c.-3/40 d.7/40
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Answer:

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8 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
3 years ago
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