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3 years ago

Pleaseee helpppp mehhhh

Mathematics

1 answer:

Tanya [424]3 years ago

6 0

Answer:

-13x^2 + 6x + 11

Step-by-step explanation:

Just subtract the numbers that are similar (i.e. -5x^2 - 8x^2)

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Find the area of a circular pond having radius Ans: 616 m 14 m.

krek1111 [17]

Step-by-step explanation:

Area of circular pond=22/7×14×14

=22/7×196

=616m

3 0

2 years ago

What is the perimeter and area of 9ft width and 8ft Long?

Sever21 [200]

Answer:

Perimeter = 34ft Area = 72ft squared

Step-by-step explanation:

The perimeter of an object is all of it's sides combined. Because you stated two values, I assumed your shape is a rectangle. So the perimeter would be:

8 + 8 + 9 + 9 = 34

The area is length times width, so:

8 x 9 = 72

Because it is area, the units are squared.

3 0

3 years ago

73.085 in standard form

jeka57 [31]

Eres una pinche pendejha huevona

Hija de tu re Phuta bomba madreh, tienes una cara de vaca y puedes venir a mamrmeh la vergha

7 0

3 years ago

14x + 54 = 11x + 96 x=

jenyasd209 [6]

Move the 11x to the left and the +54 to the right. Their signs will flip:

14x-11x = 96-54
3x = 42
x = 14

4 0

3 years ago

Read 2 more answers

The service department of a luxury car dealership conducted research on the amount of time its service technicians spend on each

mart [117]

Answer:

Probability that the mean service time is between 1 and 2 hours is 0.96764.

Step-by-step explanation:

We are given that a systematic random sample of 100 service appointments has been collected.

The 100 appointments showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

Let $\bar X$ = sample mean service time

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = average preparation time = 90 minutes

$\sigma$ = standard deviation = 140 minutes

n = sample of appointments = 100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, probability that the mean service time is between 60 and 120 minutes is given by = P(60 minutes < $\bar X$ < 120 minutes)

P(60 minutes < $\bar X$ < 120 minutes) = P( $\bar X$ < 120 min) - P( $\bar X$ $\leq$ 60 min)

P( $\bar X$ < 120 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{120-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z < 2.14) = 0.98382

P( $\bar X$ $\leq$ 60 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{60-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z $\leq$ -2.14) = 1 - P(Z < 2.14)

= 1 - 0.98382 = 0.01618

The above probability is calculated by looking at the value of x = 2.14 in the z table which has an area of 0.98382.

Therefore, P(60 min < $\bar X$ < 120 min) = 0.98382 - 0.01618 = 0.96764

7 0

3 years ago