Ask question

Ask question

All categories

tankabanditka [31]

2 years ago

6

The wait time for your local pizzeria to deliver food is normally distributed with µ = 40 minutes and σ = 3.5. What is the proba

bility the next pizza delivery will take over 47 minutes? 2.15% 13.6% 2.28% 4.56%

1 answer:

Scilla [17]2 years ago

7 0

Answer:

2.28%

Step-by-step explanation:

Given that:

µ = 40 minutes ; σ = 3.5.

x = > 47 minutes

Obtain the standardized score, Z

Zscore = (x - μ) / σ

P(x > 47) = (47 - 40) / 3.5

P(x > 47) = 7 / 3.5

P(x > 47) = 2

P(z > 2) = 0.02275

Hence, Probability that next pizza delivery takes over 47 minutes is 0.02275 = 2.275% = 2.28%

You might be interested in

HELP!!!! Find the area. Simplify your answer.<br> 552-9s<br> 3t2+t

Tems11 [23]

Answer:

A = (5s^2t - 9st) (3t + 1)

Step-by-step explanation:

(5s^2 - 9s) (3t^2 + t)

= 15s2t^2 + 5s^2t - 27st^2 - 9st

= 5s2t (3t + 1) - 9st (3t + 1)

= ( 5s^2t - 9st) (3t+1)

6 0

2 years ago

In Gil's middle school, 1/4 of the students take life science, 1/3 take physical science, 1/6 take earth science. What fraction

Mashutka [201]

Yes the second one Have a good day

4 0

2 years ago

achel invested her savings in two investment funds. The amount she invested in Fund A was 4 times as much as the amount she inve

klasskru [66]

Let A and 4A be the investment amounts.
The total profit is 3480=0.06A+0.20A=0.26A, so Fund A investment, A=3480/0.26=$13384.62, and Fund B is $53538.46.

4 0

3 years ago

Solve for the equation 2x+9=3x

Molodets [167]

Answer:

x=9

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

Paha777 [63]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

We know that,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

Using the trigonometric ratios, we get

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Hence, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

6 0

3 years ago