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3 years ago

How many times larger is .7 than .700

Mathematics

2 answers:

prohojiy [21]3 years ago

3 0

.7 & .700 are the same number

ziro4ka [17]3 years ago

3 0

Answer:

there the same

Step-by-step explanation:

.700 is bigger then .7 so add two 0s after and you can compare them then

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Each canvas bag will hold exactly 5 pounds of kings tomatoes. How many bags will be necessary to hold 5 kg worth of these same t

GarryVolchara [31]

1 kg = 2.25 pounds

5*2.25 = 11.25 pounds

11.25/5 = 2.25 bags

so you would need 3 bags

8 0

3 years ago

Read 2 more answers

The storage box shown has these dimensions.

Westkost [7]

Answer: D. 3/16 ft³

Step-by-step explanation:

The formula for finding volume is: V=hwl

v=volume
h=height
w=width
l=length

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

Given

h=3/4 ft

w=1/2 ft

l=1/2 ft

Substitute value into the formula

V=hwl

V=(3/4)(1/2)(1/2)

V=(3/8)(1/2)

V=3/16 ft³

Hope this helps!! :)

Please let me know if you have any questions

7 0

3 years ago

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force F⃗ =( 35 N )i^−( 37 N )j^ to

Trava [24]

Answer:

-189.8 J

Step-by-step explanation:

We are given that f

Force applied=F=35 Ni-37 Nj

Displacement, s=(-8.7m)i-(3.1m)j

We have to find the work done .

We know that

Work done= $F\cdot s$

Using the formula

Work done= $(35i-37j)\cdot (-8.7-3.1)$

Work done= $-304.5+114.7$

Using $i\cdot i=j\cdot j=k\cdot k=1,i\cdot j=j\cdot i=j\cdot k=k\cdot j=i\cdot k=k\cdot i=0$

Work done=-189.8J

Hence, the work done =-189.8 J

5 0

3 years ago

A circular clock face is stenciled along its edge. The radius of the clock is 12 centimeters, so the stenciling covers about

jarptica [38.1K]

Circumference = 3.14*2r
C = 3.14 * 24
C = 75.36

4 0

4 years ago

The mean score of 15 students was 82.0. (a) Another student joins, and the mean becomes 82.5. What is the new student's score? (

slega [8]

Answer:

a) The new student's score is 90.

b) Their scores are a pair of scores lower than 100 with a mean of 90.5.

Step-by-step explanation:

The mean is the sum of all scores divided by the number of students.

The mean score of 15 students was 82.0

This means that there are 15 students and the sum of the scores is 15*82 = 1230.

(a) Another student joins, and the mean becomes 82.5. What is the new student's score?

Now there are 16 students, the sum of the scores is 1230 + x and the mean is 82.5. So

$82.5 = \frac{1230 + x}{16}$

$1230 + x = 16*82.5$

$x = 1320 - 1230$

$x = 90$

The new student's score is 90.

(b) Suppose instead that two students join, taking the average from 82 to 83. What could their scores be, if they are restricted to be integer (i.e. whole) numbers?

Now there are 2 new students, so 17 in total. The mean score of the new students is 2x. The total score of the class is 1230 + 2x and the mean is 83. So

$83 = \frac{1230 + 2x}{17}$

$1230 + 2x = 1411$

$2x = 181$

$x = 90.5$

The mean of the two students scores is 90.5.

So their scores are a pair of scores lower than 100 with a mean of 90.5.

4 0

3 years ago