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Airida [17]
3 years ago
9

The numbers 38 and 41 are rounded to the nearest ten and then multiplied to estimate the product. What is the best estimate of 3

8 × 41 based on this method?
1,200
1,600
2,000
2,500


The numbers 38 and 41 are rounded to the nearest ten and then multiplied to estimate the product. What is the best estimate of 38 × 41 based on this method?

1,200
1,600
2,000
2,500




What is the product of 21 × 3?

21
24
60
63




Yusef wants to buy 20 pairs of sneakers to donate to a local shelter. Each pair costs $23. How much money will Yusef spend on all the sneakers?

$100
$200
$360
$460




The product of 42 × 31 is
.

A basketball team has 22 games this season. At each game, 95 raffle tickets are sold. How many raffle tickets are sold during all 22 games?

1,900
1,980
2,080
2,090



Carter rewrites 28 × 2 in expanded form to find the product.

28 × 2

= (20 × 2) + (8 × 2)

What is the next step to find the product?

adding the product of 20 × 2 to the product of 8 × 2
subtracting the product of 20 × 2 from the product of
8 × 2
multiplying the product of 20 × 2 by the product of 8 × 2
dividing the product of 20 × 2 by the product of 8 × 2



What is the product of 33 × 2?

31
33
60
66





What is the product?

57 × 4

228
232
240
285




Marta has 128 stamps. Kelli has 11 times as many as Marta. Marta estimates the total they have altogether by rounding each amount to the nearest ten and then multiplying.

What is the best estimate of 128 × 11 using this method?

1,200
1,300
1,408
1,508

pls hurry first one right gets brainlist
Mathematics
1 answer:
Rudik [331]3 years ago
3 0

Answer:

Jeez this is a big question

Step-by-step explanation:

1. if they are rounded to the nearest 10 that would mean that it is actually

40 * 40

40 * 40 is 1600. The answer for #1 is 1600

2. its just asking what 21 x3 is, 21x3 is 63

3. All you do for this is 20*23

20 * 23 is 460

4. 42 * 31 is 1302

5. 22 * 95 = 2090

6. You do 20x2 which is 40 and then you do 8x2 which is 16

Adding

then you have 40 + 16 = 56

Subtracting

40 - 16 = 24

Multiplying

40 * 16 = 640

Dividing

40/16 = 2.5

7. 33x2 is 66

8. 57* 4 is 228

9. 128 * 11 is 1408 but if you round the 11 to the nearest 10, that would be 10, then if you round the 128 to the nearest 10, that would be 130, and then 130 * 10 is 1300

You might be interested in
Least greatest ordering integers
morpeh [17]

1. 5, 19, 24, 46, 77, then 98


2. -62, -47, -11, 38, then 56


5 0
3 years ago
The weights of the fish in a certain lake are normally distributed with a mean of 11 lb and a standard deviation of 6. if 4 fish
Alexus [3.1K]

Since the sample standard deviation is now known, we use the z-score test. The formula is given as:

z= (X – μ) / (s / sqrt(n))

Where,

X = sample mean = 8.6 lb to 14.6 lb

μ = population mean = 11 lb

s = population standard deviation = 6

n = sample size = 4

1st: Calculating for z when x = 8.6 lb

z = (8.6 – 11) / (6 / sqrt4)

z = - 0.8

Using the standard distribution table for z:

 Probability (x = 8.6 lb) = 0.2119

 

2nd: Calculating for z when x = 14.6 lb

z = (14.6 – 11) / (6 / sqrt4)

z = 1.2

Using the standard distribution table for z:

 Probability (x = 14.6 lb) = 0.8849


Therefore the probability that the mean weight will be between 8.6 and 14.6 lb:

Probability (8.6 ≤ x ≤ 14.6 ) = 0.8849 - 0.2119

Probability (8.6 ≤ x <span>≤ 14.6 ) = 0.673                         (ANSWER)</span>

8 0
3 years ago
Solve for n.<br> n−10=5−4n <br> Enter your answer in the box.<br><br> n =
vodka [1.7K]
N-10=5-4n
n+4n=5+10
5n=15
n=3
4 0
3 years ago
21. Who is closer to Cameron? Explain.
pickupchik [31]

Problem 21

<h3>Answer:  Jamie is closer</h3>

-----------------------

Explanation:

  • A = Arthur's location = (20,35)
  • J = Jamie's location = (45,20)
  • C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

=================================================

Problem 22

<h3>Answer:  Arthur is closest to the ball</h3>

-----------------------

Explanation:

We have these key locations:

  • A = Arthur's location = (20,35)
  • J = Jamie's location = (45,20)
  • C = Cameron's location = (65,40)
  • B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

  • AB
  • JB
  • CB

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

-------------

So in summary, we have these approximate segment lengths

  • AB = 29.1547595
  • JB = 41.2310563
  • CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

3 0
3 years ago
Triangle A B C is shown. The length of B C is 6 and the length of A B is 2 StartRoot 2 EndRoot. Angle A B C is 80 degrees. Trigo
lawyer [7]

Answer:

8.4 square units

Step-by-step explanation:

See attachment for the figure.

Trigonometric area formula is Area =\frac{1}{2}\times absin(C)

Where a and b are two sides of the triangle and C is the angle between these two sides.

Given:

side a = 6 units, c = 2√2 units and ∠B = 80°

Then area of the triangle :

Area = \frac{1}{2}\times acsin(B)

       = \frac{1}{2}\times acsin(B)

       = \frac{1}{2}\times 6\times 2\sqrt{2}\times sin(80)

       = 6√2×sin80°

       = 6×(1.414)×0.9848

       = 8.36 square units

       ≈ 8.4 square units

6 0
3 years ago
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