All categories

3 years ago

Round 50.468 to the nearest thousandtha)50.000b)51.000c)51.400d)51.500

Mathematics

1 answer:

Olenka [21]3 years ago

6 0

Ok it should be B if I’m rounding that up right

You might be interested in

The ratio of the perimeters of two rectangles is 4 to 7. The perimeter of the larger rectangle is 42 inches. What is the perimet

Lelu [443]

P = 42(4/7)

p = (42)(4)/(1)(7)

p= 168/7

p= 24 inches

Answer : D , 24 inches

I hope that's help !

6 0

4 years ago

Find the values of the ratios

OleMash [197]

Answer:

perimeter: 5/8
area: 25/64

Step-by-step explanation:

You want the ratios (red to blue), of perimeter and area, given that corresponding sides have the ratio red to blue = 5 to 8.

<h3>Perimeter ratio</h3>

The ratio of perimeters is the ratio of corresponding side lengths:

red to blue = 5 / 8

<h3>Area ratio</h3>

The ratio of areas is the square of the ratio of corresponding side lengths:

red to blue = (5 to 8)² = 25 / 64

<em>Additional comment</em>

Ratios are customarily expressed in any of three equivalent forms:

5 to 8
5:8
5/8 "fraction" form

3 0

1 year ago

PLEASE HELP!!!! Given the equation y = 2x - 8, what is the slope and the y-intercept?

GaryK [48]

Y=mx+b
so m=2 and b=-8

4 0

4 years ago

Read 2 more answers

Name the three different types of proofs you saw in this lesson

KiRa [710]

Answer:

um idk but thanks for the points

8 0

3 years ago

On the last three physics exams a student scored 89, 87, and 92. What score must the student earn on the next exam to have an av

OlgaM077 [116]

Answer:

For the average to be above 90, the marks in the fourth paper should be MORE THAN 92.

Step-by-step explanation:

The score of students in last 3 exams = 89 , 87 and 92

Let us assume the marks scored by the student in last exam = k

The average of marks should be above 90.

Now, $\textrm{The average of total marks} = \frac{\textrm{Sum of all marks obtained in 4 papers}}{\textrm{ 4 papers }}$

Sum of all 4 paper's marks = 89 + 87 + 92 + k = 268 + k

Also, 90 < Average

$\implies 90 < \frac{268 + k}{4}\\ or, 360 < 268 + k\\\implies 360 - 268 < k\\or, k > 92$

Hence,for the average to be above 90, the marks in the fourth paper should be MORE THAN 92.

4 0

4 years ago

Round 50.468 to the nearest thousandtha)50.000b)51.000c)51.400d)51.500​

Round 50.468 to the nearest thousandtha)50.000b)51.000c)51.400d)51.500