All categories

3 years ago

Fnd the value of x................

Mathematics

1 answer:

lilavasa [31]3 years ago

4 0

So to solve a,
You already have the first two answers which are 65 and 90. Since a tríenle always has to equal 180 you just add 65 and 90 and the subtract that number by 180 and that’s your other missing angle.
Both triangles are identical so you just have to insert the answers from the right triangle onto the left triangle.
A is the only one that I knew how to solve, sorry

You might be interested in

Henry bought a new car for $24,000.He made a down payment of 20% and financed the balance for 10% with 24 equal monthly payments

igomit [66]

The answer It is $800.00

8 0

4 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

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