A football player runs 10.7 m/s toward the end zone. If he runs for 3.7

Han's garden has an area of 193.75

ycow [4]

Answer:

15.5

Step-by-step explanation:

The equation to find area is Length × Width= Area

we have the area and the length, 12.5 × width= 293.75

to find the width, you will have to reverse engineer and divide the area by the length instead.

193.75 ÷ 12.5 =15.5

just to double check, 15.5 × 12.5 = 193.75

the width is 15.5

3 0

3 years ago

Ibrahim likes to run a loop around the park near his house that is ⅞ mile long. There is a water fountain ½ way around the loop.

tamaranim1 [39]

Answer:

7/16 mile

Step-by-step explanation:

Distance of the loop = 7/8 mile

Distance of Water fountain = 1/2 of the Distance of the loop

= 1/2 of 7/8

Ibrahim stopped to get a drink of water at the water fountain. How far did Ibrahim run?

= 1/2 of 7/8

= 1/2 * 7/8

= (1 * 7) / (2 * 8)

= 7/16

Ibrahim ran 7/16 mile to drink water at the water fountain around the loop

8 0

3 years ago

Help w this plsss,!,!,, 20pointsss

Kay [80]

Answer:

Area = Length × width

16 = (3x +2 ) × ( x)

16 = 3x² +2x

3x² +2x -16=0

( 3x+8 ) ( x -2 ) =0

3x +8=0 —> 3x = – 8 —> x = – 8/3 —> x = – 2.6

x-2=0 —> x = 2

I hope I helped you^_^

5 0

2 years ago

A woodworking machine decreased the thickness of a board from 3/4 of in inch to 9/16 of an inch. by what number of inches (i) di

nataly862011 [7]

The thickness of the board was decreased from 3 / 4 inches to 9 / 16 inches.

Therefore,

$\text{ the decrement = }\frac{3}{4}-\frac{9}{16}=\frac{12-9}{16}=\frac{3}{16}\text{ inches}$

So the board's thickness in inches decreased by

$\frac{3}{16}$

6 0

1 year ago

6. (4.2.12) Of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired. a. F

nata0808 [166]

Answer:

(a) Probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.

Step-by-step explanation:

We are given that of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired.

Let Probability that item are defective = P(D) = 0.20

Also, R = event of item being repaired

Probability of items being repaired from the given defective items = P(R/D) = 0.60

So, Probability of items not being repaired from the given defective items = P(R'/D) = 1 - P(R/D) = 1 - 0.60 = 0.40

(a) Probability that a randomly chosen item is defective and cannot be repaired = Probability of items being defective $\times$ Probability of items not being repaired from the given defective items

= 0.20 $\times$ 0.40 = 0.08 or 8%

So, probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Now we have to find the probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 20 items

r = number of success = exactly 2

p = probability of success which in our question is % of randomly

chosen item to be defective and cannot be repaired, i.e; 8%

LET X = Number of items that are defective and cannot be repaired

So, it means X ~ $Binom(n=20, p=0.08)$

Now, Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is given by = P(X = 2)

P(X = 2) = $\binom{20}{2} \times 0.08^{2} \times (1-0.08)^{20-2}$

= $190 \times 0.08^{2} \times 0.92^{18}$

= 0.2711

Therefore, probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.

6 0

3 years ago