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2 years ago

A yard contains 16 ounces. 9 ounces is what fraction of a pound?

Mathematics

2 answers:

andreev551 [17]2 years ago

7 0

Answer:

1.3

Step-by-step explanation:

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Svetradugi [14.3K]2 years ago

4 0

9514 1404 393

Answer:

9/16

Step-by-step explanation:

We presume an error in the problem statement. A yard is a measure of distance, not weight.

A pound contains 16 ounces. Then 9 ounces is 9/16 of a pound.

As a decimal fraction, that is 0.5625 lb.

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SOCCER PRACTICE BEGAN AT 2:30 P.M 3:30 P.M THROUGH WHAT FRACTION OF A CIRCLE DID THE MINUTE HAND TURN HOW MANY DEGREES DID THE M

MrMuchimi

Answer:

1 /1 ; 360°

Step-by-step explanation:

Start time = 2:30 pm

Stop time = 3:30 pm

Minute hand makes a complete revolution per hour ;

This means that the minutes hand revolves round the whole circle in one hour, hence fraction of the circle covered by the minute hand between 2:30 pm - 3:30 pm is 1/1 = 1

The angle turned by the minutes hand :

1 complete revolution = 360°

A circle covers 360°

1 /1 fraction of a circle = 1/1 * 360° = 360°

Hence, angle turned by the minute hand = 360°

5 0

3 years ago

5x-3 and 2x+ 25, find the value of x

andreev551 [17]

If they are equal then,
5x-3=2x+25
3x=22
X=7 +1/3=7.(3)

3 0

2 years ago

Read 2 more answers

An inequality is shown. 27/7n > 4/3 Select all the values of n that make the inequality true.

ankoles [38]

Answer:

Step-by-step explanation:

<h2>Hope it help & study well</h2>

6 0

2 years ago

What is the value of r (3,4), (-1,r),m=-3/4

Llana [10]

To find the value of r

We will follow the steps below

Usinf the equation of a slope:

slope(m) =

$m=\frac{y_2-y_1}{x_2-x_1}$

from the question given;

m= -3/4

x₁ = 3

y₁=4

x₂=-1

y₂=r

substituting the values into the slope formula

$\frac{-3}{4}\text{ = }\frac{r-4}{-1-3}$

we can now simplify and then solve for r

$\frac{-3}{4}\text{ = }\frac{r-4}{-4}$

cross-multiply

-3 x -4 = 4(r-4)

12 = 4(r-4)

Divide both-side of the equation by 4

$\frac{12}{4}\text{ = }\frac{4(r-4)}{4}$

3 = r - 4

add 4 to both-side of the equation

3+4 = r-4+4

7 = r

r=7

Therefore the value of r is 7

4 0

1 year ago

Let X be the number of the cars being repaired at a repair shop. We have the following information:

worty [1.4K]

Answer:

(a) Sample Space

$S = \{0,1,2,3\}$

(b) PMF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

(c) CDF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

$S = \{0,1,2,3\}$

Solving (b): The PMF

From the question, we have:

$P(2) = P(1)$

$P(0) = P(3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$ can be represented as:

$P(1) + P(2) = 0.5[P(0) + P(3)]$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(1) + P(1) = 0.5[P(0) + P(0)]$

$2P(1) = 0.5[2P(0)]$

$2P(1) = P(0)$

$P(0)= 2P(1)$

Also note that:

$P(0) + P(1) + P(2) + P(3) = 1$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(0) + P(1) + P(1) + P(0) = 1$

$2P(1) + 2P(0) = 1$

Substitute $P(0)= 2P(1)$

$2P(1) + 2*2P(1) = 1$

$2P(1) + 4P(1) = 1$

$6P(1) = 1$

Solve for P(1)

$P(1) = \frac{1}{6}$

To calculate others, we have:

$P(2) = P(1)$

$P(2) = P(1) = \frac{1}{6}$

$P(0)= 2P(1)$

$P(0) =2 * \frac{1}{6}$ $P(0) =\frac{1}{3}$

$P(3) = P(0) =\frac{1}{3}$

Hence, the PMF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

<em>See attachment (1) for histogram</em>

Solving (c): The CDF ; F(x)

This is calculated as:

$F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)$

For x = 0;

We have:

$P(X \le 0) = P(0)$

$P(X \le 0) = 1/3$

For x = 1

$P(X \le 1) = P(0) + P(1)$

$P(X \le 1) = 1/3 + 1/6$

$P(X \le 1) = 1/2$

For x = 2

$P(X \le 2) = P(0) + P(1) + P(2)$

$P(X \le 2) = 1/3 + 1/6 + 1/6$

$P(X \le 2) = 2/3$

For x = 3

$P(X \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3$

$P(X \le 3) = 1$

Hence, the CDF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

<em>See attachment (2) for histogram</em>

6 0

3 years ago