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2 years ago

1537 × 242 showing working

Mathematics

2 answers:

Veseljchak [2.6K]2 years ago

6 0

1537
x242

3074
61480
307400

371,954 is the answer

Dima020 [189]2 years ago

3 0

Number sentence: 1537 x 242 = ?

Answer: 371954

Work shown:

First i multiplied 1000 by 242 and got 242000

Then i multiplied 500 by 242 and got 121000

Then i multiplied 30 by 242 and got 7260

Then i multiplied 7 by 242 and got 1694

And finally i added them up like 242000 + 121000 + 7260 + 1694 and that got me 371954.

(Let me know if i got my additions wrong)

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There are 345 students at a college who have taken a course in calculus, 212 who have taken a course in discrete mathematics, an

ollegr [7]

Answer:

369 students have taken a course in either calculus or discrete mathematics

Step-by-step explanation:

I am going to build the Venn's diagram of these values.

I am going to say that:

A is the number of students who have taken a course in calculus.

B is the number of students who have taken a course in discrete mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the number of students who have taken a course in calculus but not in discrete mathematics and $A \cap B$ is the number of students who have taken a course in both calculus and discrete mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

188 who have taken courses in both calculus and discrete mathematics.

This means that $A \cap B = 188$

212 who have taken a course in discrete mathematics

This means that $B = 212$

345 students at a college who have taken a course in calculus

This means that $A = 345$

How many students have taken a course in either calculus or discrete mathematics

$(A \cup B) = A + B - (A \cap B) = 345 + 212 - 188 = 369$

369 students have taken a course in either calculus or discrete mathematics

4 0

1 year ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

3 0

1 year ago

Which statement best summarizes the central idea of “Applications of Newton’s Laws”? Newton’s laws can explain the forces that o

nadya68 [22]

Answer: Newton’s laws can explain the forces that occur between objects every day.

Newton's Laws are known as the "Three Laws of Motion." They address inertia, force, and gravity. Because these are all phenomenons that occur every day, that is the correct answer.

5 0

1 year ago

Read 2 more answers

the distribution of scores on a recent test closely followed a normal distribution wotb a mean of 22 and a standard deviation of

soldi70 [24.7K]

Answer:

1) 22.66%

2) 20

Step-by-step explanation:

The scores of a test are normally distributed.

Mean of the test scores = u = 22

Standard Deviation = $\sigma$ = 4

Part 1) Proportion of students who scored atleast 25 points

Since, the test scores are normally distributed we can use z scores to find this proportion.

We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:

P(X ≥ 25)

We can convert this value to z score and use z table to find the required proportion.

The formula to calculate the z score is:

$z=\frac{x-u}{\sigma}$

Using the values, we get:

$z=\frac{25-22}{4}=0.75$

So,

P(X ≥ 25) is equivalent to P(z ≥ 0.75)

Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266

Since,

P(X ≥ 25) = P(z ≥ 0.75), we can conclude:

The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%

Part 2) 31st percentile of the test scores

31st percentile means 31%(0.31) of the students have scores less than this value.

This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.

Using the z table, the z score for 31st percentile comes out to be:

z = -0.496

Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.

Using the values, we get:

$-0.496=\frac{x-22}{4}\\\\ x=4(-0.496) + 22\\\\ x=20.02\\\\ x \approx 20$