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Leni [432]
3 years ago
7

Round 62.85 to the nearest tenth

Mathematics
2 answers:
Fiesta28 [93]3 years ago
5 0

Answer:

62.9

Step-by-step explanation:

Follow the rounding rules.

Remember the tenths place is one place after the decimal point.

Dimas [21]3 years ago
5 0
The answer is 62.9. Hope this helps!!
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Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best po
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Answer:

Step-by-step explanation:

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3 0
2 years ago
The figure on the left represents a scale drawing of the figure on the right. What is the scale?
docker41 [41]

Answer:

\frac{1}{90}

Step-by-step explanation:

Before calculating the scale we require the dimensions to be in the same units.

Using the conversion

1 yard = 3 ft and

1 foot = 12 inches, then

5 yards = 5 × 3 × 12 = 180 inches

The scale is then

2 in : 180 in ← divide both quantities by 2

= 1 : 90

= \frac{1}{90}

4 0
4 years ago
The perimeter of a rectangular town is 32 miles.its width is 5 miles find the length
son4ous [18]

The perimeter of a rectangle is expressed as:

P = perimeter

l = length

w = width

P = 2(l + w)

Plug in our values to the formula mentioned above

32 m = 2(l + 5).

Start by dividing each side by 2.

16 = l + 5.

Subtract 5 from each side.

11 = l.

The length of your town is 11 miles

8 0
4 years ago
Laura wrote and solved the following expression to find the total number of beads Lina has if there are 6 beads in each packet.
bulgar [2K]
<h3>The number of beads lina has is 31</h3><h3>Laura made a mistake by adding the constant 7 and variable 4b</h3>

<em><u>Solution:</u></em>

Given that,

Laura wrote and solved the following expression to find the total number of beads Lina has

There are 6 beads in each packet

7+4b = 11b

= 11(6)

=66

From given,

Lina has a total of 7 + 4b beads

Given that, There are 6 beads in each packet

Substitute b = 6

7 + 4(6)

Simplify

7 + 24

Add

31

Thus, she has a total of 31 beads and not 66 beads

Laura made a mistake by adding the constant 7 and variable 4b

But a constant and variable cannot be added

4 0
4 years ago
A) If a:b = 2:5 and b:c = 3:4, find (i) a:c(ii) a:b:c
kobusy [5.1K]

Answer:

A i. a:c=3:10

ii. a:b:c=2:5:10

B i. x:z=2:5

ii. x:y:z=2:4:5

Step-by-step explanation:

A.) If a:b = 2:5 and b:c = 3:4, find (i) a:c(ii) a:b:c

a:b=a/b=2/5

b:c=b/c=3/4

a/b*b/c=a/c

2/5*3/4=a/c

6/20=a/c

3/10=a/c

Therefore, a:c=3:10

a:b:c

a:b=2:5

b:c=3:4

b is common to both ratios

The value of b in the first ratio is 5 and b is 3 in the second ratio

Lets take the LCM of both values

LCM of 5 and 3=15

So, we will change the value of b in the first ratio and second ratio to 15

By doing this, we will multiply the whole first ratio by 3

We have, 6:15

We multiply the whole second ratio by 5

We have, 15:20

Therefore a:b:c=6:15:20

=2:5:10

B. If x:y = 1:2 and y:z = 4:5,

x:y=x/y=1:2

y:z=y/z=4:5

x/y*y/z=x/z

1/2*4/5=x/z

4/10=x/z

2/5=x/z

Therefore, x:z=2:5

x:y:z

x:y=1:2

y:z=4:5

y is common to both ratio

Take the LCM of y values in both ratio

LCM of 2 and 4 =4

So,we will change the value of y in the first and second ratio to 4

By doing this, we will multiply the whole first ratio by 2

We have, 2:4

We will also multiply the whole second ratio by 1

We have, 4:5

Therefore, x:y:z=2:4:5

8 0
3 years ago
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