The length of a rectangular garden is 8 feet longer than its width. If the garden's perimeter is 184 feet,

Another bowler, Ryan Shafer, has a mean absolute deviation of 5. Based on the mean absolute deviation of Norm Duke, who has been

Wewaii [24]

Answer:

Ryan is a more consistent performer

Step-by-step explanation:

The complete question is

HEL 1. Shown are the final scores for Norm Duke in the past eight bowling tournaments.

299, 281, 285, 269, 280, 269, 286, 287

Part A: What is the mean absolute deviation of Norm Duke’s scores?

A.

7.50

B.

7.00

C.

7.25

D.

6.75

Part B: Based on the mean absolute deviation from Part A, who has been the more consistent bowler if Ryan Shafer has a mean absolute deviation of 7?

Solution

First we will find the mean deviation of Duke is 7.5

The standard deviation of Ryan is 7.0. This means that the average variation in case of Duke is higher than that of average variation of Ryan.

Hence, Ryan has been more consistent performer

4 0

2 years ago

Three consecutive integers have a sum of 168. What is the value of the largest integer

Svet_ta [14]

Largest integer: 57

x + x + 1 + x + 2 = 168
3x + 3 = 168
-3 -3
3x = 165
/3 /3
x = 55

x = smallest integer = 55
x + 1 = middle integer = 55 + 1 = 56
x + 2 = largest integer = 55 + 2 = 57

6 0

3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

At a clearance sale,Jill got 70% off a silver bracelet.If she paid the cashier $27 ,what was the original price of the bracelet

mart [117]

Answer:

$38.57

Step-by-step explanation:

Make a proportion.

$\frac{27}{x} = \frac{70}{100}$

Cross multiply.

27 times 100 = 2700

2700 divided by 70 = 38.57142857

Round the result to the hundredths place.

38.57142857 ≈ 38.57

**Hope this helps!

4 0

3 years ago

The distance from around the Earth along a given latitude can be found using the formula C=2∏r cos L , where r is the radius of

guajiro [1.7K]

Answer:

The distance reduces to 0 as you go from 0° to 90°

Step-by-step explanation:

The question requires you to find the distance using different values of L and check the trend of the values.

Given C=2×pi×r×cos L where L is the latitude in ° and r is the radius in miles then;

Taking r=3960 and L=0° ,

C=2× $\pi$ ×3960×cos 0°

C=2× $\pi$ ×3960×1

C=7380 $\pi$

Taking L=45° and r=3960 then;

C= 2× $\pi$ ×3960×cos 45°

C=5600.28 $\pi$

Taking L=60° and r=3960 then;

C=2× $\pi$ ×3960×cos 60°

C=3960 $\pi$

Taking L=90° and r=3960 then;

C=2× $\pi$ ×3960×cos 90°

C=2× $\pi$ ×3960×0

C=0

Conclusion

The values of the distance from around the Earth along a given latitude decreases with increase in the value of L when r is constant

7 0

3 years ago

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