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torisob [31]
2 years ago
13

The weight of a soccer ball is normally distributed with a mean of 21 oz and a standard deviation of 3 oz. Suppose 1000 differen

t soccer balls are in a warehouse. About how many soccer balls weigh more than 24 oz
A. 40

B. 80

C. 160

D. 200
Mathematics
2 answers:
horsena [70]2 years ago
8 0

Answer:

The correct option is C.

Step-by-step explanation:

Given information: The weight of a soccer ball is normally distributed with a mean of 21 oz and a standard deviation of 3 oz. Total number of soccer balls is 1000.

We have to find the number of soccer balls having weight more than 24 oz.

Probability of the ball having weight more than 24 oz is

P(x>24)=P(\frac{x-\mu}{\sigma}>\frac{24-21}{3})

P(x>24)=P(z>1)

P(x>24)=1-P(z\leq 1)

P(x>24)=1-P(z\leq 1)                  (Using standard normal table)

P(x>24)=1-0.8413

P(x>24)=0.1587

The number of soccer balls having weight more than 24 oz is

1000\times P(x>24)=1000\times 0.1587\Rightarrow 158.7\approx 160

The number of soccer balls having weight more than 24 oz is about 160.

Therefore the correct option is C.

Novosadov [1.4K]2 years ago
3 0

A weight of 24 oz is 1 standard deviation above the mean 21 oz. The empirical rule says approximately 68% of soccer balls have weights within 1 standard deviation (i.e. between 18 and 24 oz), so the remaining 32% of balls have weights below 18 oz or above 24 oz. The balls' weights are normally distributed, so the percentage of balls with weight less than 18 oz is the same as the percentages of balls with weight greater than 24 oz, which means about 16% of balls have weight greater than 24 oz. 16% of 1000 is 160.

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You are designing a miniature golf course and need to calculate the surface area and volume of many of the objects that will be
laiz [17]
a. To solve the first part, we are going to use the formula for the surface area of a sphere: A=4 \pi r^2
where
A is the surface area of the sphere
r is the radius of the sphere
We know from our problem that r=5ft; so lets replace that value in our formula:
A=4 \pi (5ft)^2
A=314.16ft^2

To solve the second part, we are going to use the formula for the volume of a sphere: V= \frac{4}{3}  \pi r^3
Where
V is the volume of the sphere
r is the radius 
We know form our problem that r=5ft, so lets replace that in our formula:
V= \frac{4}{3}  \pi (5ft)^3
V=523.6ft^3

We can conclude that the surface area of the sphere is 314.16 square feet and its volume is 523.6 cubic feet.

b. To solve the first part, we are going to use the formula for the surface area of a square pyramid: A=a^2+2a \sqrt{ \frac{a^2}{4} +h^2}
where
A is the surface area
a is the measure of the base
h is the height of the pyramid 
We know form our problem that a=8ft and h=12ft, so lets replace those value sin our formula:
A=(8ft)^2+2(8ft) \sqrt{ \frac{(8ft)^2}{4} +(12ft)^2}
A=266.39ft^2

To solve the second part, we are going to use the formula for the volume of a square pyramid: V=a^2 \frac{h}{3}
where
V is the volume 
a is the measure of the base
h is the height of the pyramid
We know form our problem that a=8ft and h=12ft, so lets replace those value sin our formula:
V=(8ft)^2 \frac{(12ft)}{3}
V=256ft^3

We can conclude that the surface area of our pyramid is 266.39 square feet and its volume is 256 cubic feet.

c. To solve the first part, we are going to use the formula for the surface area of a circular cone: A= \pi r(r+ \sqrt{h^2+r^2}
where
A is the surface area
r is the radius of the circular base
h is the height of the cone
We know form our problem that r=5ft and h=8ft, so lets replace those values in our formula:
A= \pi (5ft)[(5ft)+ \sqrt{(8ft)^2+(5ft)^2}]
A=226.73ft^2

To solve the second part, we are going to use the formula for the volume os a circular cone: V= \pi r^2 \frac{h}{3}
where
V is the volume
r is the radius of the circular base
h is the height of the cone 
We know form our problem that r=5ft and h=8ft, so lets replace those values in our formula:
V= \pi (5ft)^2 \frac{(8ft)}{3}
V=209.44ft^3

We can conclude that the surface area of our cone is 226.73 square feet and its surface area is 209.44 cubic feet.

d. To solve the first part, we are going to use the formula for the surface area of a rectangular prism: A=2(wl+hl+hw)
where
A is the surface area
w is the width
l is the length 
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We know from our problem that w=6ft, l=10ft, and h=16ft, so lets replace those values in our formula:
A=2[(6ft)(10ft)+(16ft)(10ft)+(16ft)(6ft)]
A=632ft^2

To solve the second part, we are going to use the formula for the volume of a rectangular prism: V=whl
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V is the volume 
w is the width
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We know from our problem that w=6ft, l=10ft, and h=16ft, so lets replace those values in our formula:
V=(6ft)(16ft)(10ft)
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We can conclude that the surface area of our solid is 632 square feet and its volume is 960 cubic feet.

e.  Remember that a face of a polygon is a side of polygon.
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Total faces: 5 + 1 + 6 = 12 faces

<span>We can conclude that there are 12 faces in on the four geometric shapes on the holes.
</span>
f. Remember that an edge is a line segment on the boundary of the polygon.
   - A sphere has no edges.
   - A cone has no edges.
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Total edges: 8 + 20 = 20 edges

Since we have 20 edges in total, we can conclude that your boss will need 20 brackets on the four shapes.

g. Remember that the vertices are the corner points of a polygon.
   - A sphere has no vertices.
   - A cone has no vertices.
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   - A rectangular prism has 8 vertices.
Total vertices: 5 + 8 = 13 vertices

We can conclude that there are 0 vertices for the sphere and the cone; there are 5 vertices for the pyramid, and there are are 8 vertices for the solid (rectangular prism). We can also conclude that your boss will need 13 brackets for the vertices of the four figures.

7 0
2 years ago
Bilet ulgowy do cyrku jesto 10 zł tańszy od biletu normalnego za 3 bilety normalne i 2 ulgowe zapłacono 160 zł ile należało by z
kipiarov [429]

Answer:

Kwota zapłacona za 2 bilety normalne i 1 bilet ulgowy wynosi 98 złotych

Step-by-step explanation:

Po pierwsze, zadeklarujmy zmienne. Niech kwota zapłacona za normalne bilety wyniesie n złotych, a kwota zapłacona za bilety zniżkowe będzie d złotych

Od pierwszej części pytania powiedziano nam, że kwota za bilet ulgowy jest o 10 złotych mniejsza niż w przypadku biletu normalnego.

Stąd matematycznie n = d + 10 ......... (i)

Mówi się nam również, że kwota zapłacona za 3 normalne bilety i 2 bilety ze zniżką wynosi 160 złotych

Matematycznie można to zapisać jako;

3n + 2d = 160 .................... (ii)

Teraz, aby uzyskać kwotę zapłaconą za dwa bilety normalne i jeden bilet zniżkowy, musimy najpierw znaleźć kwotę zapłaconą za każdy bilet, korzystając z dwóch powyższych równań. Teraz zamień (i) na (ii)

mamy; 3 (d + 10) + 2d = 160

3d + 30 + 2d = 160

5d + 30 = 160

5d + 160-30

5d = 130

d = 130/5

d = 26 złotych

Kwota zapłacona za normalny bilet wynosi zatem = 26 + 10 = 36 złotych

Znajdując kwotę zapłaconą za 2 bilety normalne i 1 bilet zniżkowy, mamy 2 (36) + 26 = 72 + 26 = 98 złotych

3 0
1 year ago
Michaela pays her cell phone service provider $49.95 per month for 500 minutes. any additional minutes used cost $0.15 each. in
lilavasa [31]
75 extra minutes because $61.20-$49.95=$11.25 and $11.25/$0.15=75.
7 0
2 years ago
Read 2 more answers
What value of x would make KM ∥ JN?
stepan [7]

Answer:

x = 10

Step-by-step explanation:

By the converse of the side-splitter theorem, if \dfrac{JK}{KL}= \dfrac{NM}{ML} , then KM ∥ JN.

Substitute the expressions into the proportion:

\dfrac{x-5}{x}= \dfrac{x-3}{x+4}

Cross-multiply:

(x – 5)(x+4) = x(x – 3).

Distribute:

x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3).

Multiply and simplify:

x^2 +4x- 5x - 20 = x^2-3x\\-x-20=-3x\\$Collect like terms$\\-x+3x=20\\2x=20\\$Divide both sides by 2\\x=10

Solve for x: x = 10

Therefore, the value of x that would make KM parallel to JN is 10.

4 0
1 year ago
Use the position function s(t) = −16t² + 400, which gives the height (in feet) of an object that has fallen for t seconds from a
skad [1K]

Answer:

160m/s

Step-by-step explanation:

The object can hit the ground when t = a; meaning that s(a) = s(t) = 0

So, 0 = -16a² + 400

16a² = 400

a² = 25

a = √25

a = 5 (positive 5 only because that's the only physical solution)

The instantaneous velocity is

v(a) = lim(t->a) [s(t) - s(a)]/[t-a)

Where s(t) = -16t² + 400

and s(a) = -16a² + 400

v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)

v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)

v(a) = lim (t->a) -16(t² - a²)(t-a)

v(a) = -16lim t->a (t²-a²)(t-a)

v(a) = -16lim t->a (t-a)(t+a)/(t-a)

v(a) = -16lim t->a (t+a)

But a = t

So, we have

v(a) = -16lim t->a 2a

v(a) = -32lim t->a (a)

v(a) = -32 * 5

v(a) = -160

Velocity = 160m/s

7 0
1 year ago
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