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2 years ago

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Kimo's car needed work. The mechanic charged him $140 for parts plus $48 per hour for labor. If the bill totaled $260, how many

hours of labor were required?

1 answer:

BlackZzzverrR [31]2 years ago

4 0

Hi!

We will find the hours of labor, x, like this:

140 + 48 * x = 260

48 * x = 260 - 140

48 * x = 120

x = 120/48

x = 2,5 hours of labor were required

Hope this helps!

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Answer:

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D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.

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2 years ago

5y-6=3y-20 how do i solve this problem

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2 years ago

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A circle with radius of \greenD{5\,\text{cm}}5cmstart color #1fab54, 5, start text, c, m, end text, end color #1fab54 sits insid

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Answer:

The area of the shaded region is 42.50 cm².

Step-by-step explanation:

Consider the figure below.

The radius of the circle is, <em>r</em> = 5 cm.

The sides of the rectangle are:

<em>l</em> = 11 cm

<em>b</em> = 11 cm.

Compute the area of the shaded region as follows:

Area of the shaded region = Area of rectangle - Area of circle

$=[l\times b]-[\pi r^{2}]\\\\=[11\times 11]-[3.14\times 5\times 5]\\\\=121-78.50\\\\=42.50$

Thus, the area of the shaded region is 42.50 cm².

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2 years ago

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You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl

gtnhenbr [62]

Answer:

maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

Step-by-step explanation:

We know that area of circle is given by

$A=\pi \times (radius)^{2}$

For circle with radius 'r' we have

$A_{1}=\pi \times (r)^{2}$

For circle with radius 'R' we have

$A_{2}=\pi \times (R)^{2}$

Now according to given condition we have

$A_{1}-A_{2}\leq \frac{5}{\pi }$

$\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}$

Thus maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

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2 years ago

in a certain class, 22 pupils take one or more of chemistry, economic, government. 12 take economic (E), 8 take government (G) a

mash [69]

Answer:

ai) n(E⋂C) = ∅ = null

n(E⋂G) = 4

aii) see attachment

bi) n(C⋂G) = x = 1

bii) n(G) only = 3

Step-by-step explanation:

Let chemistry = C

Economic = E

Government = G

n(E) = 12

n(G) = 8

n(C) = 7

ai) number of pupils for economics and chemistry = 0

number of pupils for economics and government = 4

The set notation for both:

n(E⋂C) = ∅ = null

n(E⋂G) = 4

aii) find attached the Venn diagram

bi) n(C⋂G) = ?

Let number of n(C⋂G) = x

From the Venn diagram

n(C) only = 12-4 = 8

n(G) only = 8-(4+x) = 4-x

n(E) only = 7-x

n(E⋂C⋂G) = 0

n(E⋂C) = 0

n(E⋂G) = 4

Total: 8+ 4-x + 7-x + x + 0+0+4 = 22

23 -x = 22

23-22 = x

x = 1

n(C⋂G) = x = 1

Number of pupils that take both chemistry and government = 1

(bii) government only = n(G) only = 4-x

n(G) only = 4-1 = 3

Number of students that take government only = 3

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2 years ago