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3 years ago

What is 17.5% as a decimal?

Mathematics

1 answer:

Andreyy893 years ago

3 0

17.5% is the same as 17.5/100 this is equal to 0.175 as a decimal

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The bear population increases at a rate of 2% per year. There are

Margaret [11]

Answer:

The answer would be P(x) = 1573*(1.02)^x

8 0

3 years ago

Mhanifa please help with this

Fed [463]

Answer:

<u>Use cross-multiplication to solve these</u>

1) 6/2 = 4/p ⇒ 6p = 4*2 ⇒ p = 8/6 ⇒ p = 4/3
2) 4/k = 8/2 ⇒ 8k = 2*4 ⇒ k = 8/8 ⇒ k = 1
3) n/4 = 8/7 ⇒ n = 4*8/7 ⇒ n = 32/7
4) 5/3 = x/4 ⇒ x = 4*5/3 ⇒ x = 20/3

8 0

3 years ago

in the figure below trianglePQM and triangleQRP are right triangles. the measure of lineQM is 6 and the measure of lineQP is 8.

Kipish [7]

Answer:

Option 4 is correct. The length of PR is 6.4 units.

Step-by-step explanation:

From the given figure it is noticed that the triangle PQR and triangle MQR.

Let the length of PR be x.

Pythagoras formula

$hypotenuse^2=base^2+perpendicular^2$

Use pythagoras formula for triangle PQM.

$PM^2=QM^2+PQ^2$

$PM^2=(6)^2+(8)^2$

$PM^2=36+64$

$PM^2=100$

$PM=10$

The value of PM is 10. The length of PR is x, so the length of MR is (10-x).

Use pythagoras formula for triangle PQR.

$PQ^2=QR^2+PR^2$

$(8)^2=QR^2+x^2$

$64-x^2=QR^2$ .....(1)

Use pythagoras formula for triangle MQR.

$MQ^2=QR^2+MR^2$

$(6)^2=QR^2+(10-x)^2$

$36=QR^2+x^2-20x+100$

$36-x^2+20x-100=QR^2$ .... (2)

From equation (1) and (2) we get

$36-x^2+20x-100=64-x^2$

$20x-64=64$

$20x=128$

$x=6.4$

Therefore length of PR is 6.4 units and option 4 is correct.

3 0

3 years ago

If 3 tons of flower bed mulch cost $480, what is the price per pound?

Goshia [24]

1 ton = 2,000 pounds
3 tons = 6,000 pounds
$480/6,000 = 0.08
per pound is $0.08

6 0

3 years ago

Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (

IrinaVladis [17]

Answer:

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14 and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = $\dfrac{9+9}{2}$

median = $\dfrac{18}{2}$

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = $\dfrac{12+13}{2}$

Median = $\dfrac{25}{2}$

Median = 12.5

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = Q3 - Q1

The interquartile range = 12.5 - 9

The interquartile range = 3.5

7 0

3 years ago