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2 years ago

Why do we buy pre bc i dont want it

Mathematics

1 answer:

sattari [20]2 years ago

8 0

Idek man I'm just here for the points

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Help with this question please!!

kkurt [141]

Answer:

The second one, 160

Step-by-step explanation:

3+3+5+8=19, and 380/19=20 so each unit is 20cm

8 is the largest side --> 8*20 = 160

3 0

3 years ago

Can someone plz help me?

Sav [38]

Answer:

The first graph

Step-by-step explanation:

-2a - 5 > 3

-2a > 8

a < -4 (when divided by a negative number, the inequality sign will be flipped)

Therefor the first graph shows the solutions

4 0

4 years ago

I need help plzz someone help meee :)

Eddi Din [679]

Answer:

Table 1 = Proportional

y is 7 times x

Table 2 = Not proportional

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

George drives at 45mi/h. write two expressions for the number of miles George travels in h hours

GaryK [48]

45h=miles in h hours

7 0

3 years ago

Read 2 more answers

The distribution of passenger vehicle speeds traveling on a certain freeway in California is nearly normal with a mean of 71.5 m

Zinaida [17]

Answer:

$G = 0.007523$

The number of cars the highway patrol officer would watch before a car that is seen is $E(X) = 1.6027$

The standard deviation is $s = 0.9829$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 71.5 \ miles/hour$

The standard deviation is $\sigma = 4.75 \ miles/hour$

The speed limit is $x = 70 \ miles /hour$

Generally the probability of getting a car that is moving with speed greater than the speed limit is mathematically represented as

$p =P(X > x ) = P(X > 70) = P(\frac{X - \mu }{\sigma } > \frac{70 - 71.5 }{4.75})$

=> $p= P(X > 70) = P(\frac{X - \mu }{\sigma } > \frac{70 - 71.5 }{4.75})$

=> $p= P(X > 70) = P(\frac{X - \mu }{\sigma } > -0.31579 )$

Here

$\frac{X - \mu }{\sigma } = Z(The \ standardized \ value \ of X )$

=> $p= P(X > 70) = P(Z > -0.31579 )$

From the z-table

$p = P(Z > -0.31579 ) = 0.62392$

$p = P(X > 70) = 0.62392$

Generally the probability of getting a car that is not moving with speed greater than the speed limit is mathematically represented as

$q = 1 - p$

=> $q = 1 - 0.62392$

=> $q = 0.37608$

Generally the probability of getting 5 cars that are not speeding is mathematically represented as

$G = q^5$

=> $G = (0.37608)^5$

=> $G = 0.007523$

Generally the number of cars that the highway patrol officer is expected to watch until the first car that is speeding is gotten is mathematically represented as

$E(X) = \frac{1}{p}$

=> $E(X) = \frac{1}{0.62392}$

=> $E(X) = 1.6027$

Generally the standard deviation is mathematically represented as

$s = \sqrt{\frac{1 - p }{ p^2} }$

=> $s = \sqrt{\frac{1 -0.62392 }{ (0.62392)^2} }$

=> $s = 0.9829$

8 0

3 years ago