2x^2+7x-k=0 find all the values of k

4(y – 3) = y – pls help me on this question

AURORKA [14]

Answer:

y=4

Step-by-step explanation:

7 0

3 years ago

the angle of elevation of the top of a tower from a point 42m away from the base on level ground is 36 find the height of the to

Tcecarenko [31]

Answer:

30.51 meters

Step-by-step explanation:

Given that:

The distance from the point to the base of the tower = 42 m, the angle of elevation = 36°.

According to sine rule if a,b,c are the sides of a triangle and its respective opposite angles are A, B, C. Therefore:

$\frac{a}{sin(A)} =\frac{b}{sin(B)}=\frac{c}{sin(C)}$

Let the height of the tower be a and the angle opposite the height be A = angle of elevation = 36°

Also let the distance from the point to the base of the tower be b = 42 m, and the angle opposite the base of the tower be B

To find B, since the angle between the height of the tower and the base is 90°, we use:

B + 36° + 90° = 180° (sum of angles in a triangle)

B + 126 = 180

B = 180 - 126

B = 54°

Therefore using sine rule:

$\frac{a}{sin(A)} =\frac{b}{sin(B)}\\\\\frac{a}{sin(36)}=\frac{42}{sin(54)}\\\\ a=\frac{42*sin(36)}{sin(54)}\\ \\a=30.51\ meters$

The height of the tower is 30.51 meters

6 0

3 years ago

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c) The probability that a randomly selected woman shop less than 5 hours online is 0.9922.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$