If you earn $15 an hour, what will your pay be (before taxes) if you work 27 hours in one week

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it

brilliants [131]

Answer:

(a) 11.25 and 1.68

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

where, n = number of trials (samples) taken = 15

r = number of success

p = probability of success which in our question is % of cases in

which customer problems are resolved on the same day, i.e.;75%

So, here X ~ $Binom(n=15,p=0.75)$

(a) Expected number of problems to be resolved today = E(X)

E(X) = $\mu$ = n * p = 15 * 0.75 = 11.25

Standard deviation = $\sigma$ = $\sqrt{n*p*(1-p)}$ = $\sqrt{15*0.75*(1-0.75)}$ = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

P(X = 10) = $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}$

= $3003*0.75^{10} *0.25^{5}$ = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

= $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}$

= $3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4}$ = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

= $\binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}$

= $1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}$

= 0.6865

3 0

3 years ago

0+oi, 4+0iand 0+3i the area and the length

Nuetrik [128]

What's the question you're asking?

5 0

3 years ago

Hey! Can someone check my answers?

Andrew [12]

The greatest common factor is the biggest number taken from the values.

Q1. The answer is <span>A. 5y^6
</span>
$20 y^{9} +5 y^{6}= 4*5 y^{9}+5 y^{6}$
Since $x^{a}* x^{b} =x^{a+b}$ , then $x^{9}= x^{3}* x^{6}$

Back to our expression:
$4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+ 5y ^{6}*1=5 y^{6} (4y ^{3} +1)$
The greatest common factor is thus $5 y^{6}$

Q2. The answer is <span>D. 12xy^2
</span>
$12x y^{5}+60 x^{4} y^{2} -24 x^{3} y^{3}=12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}$
We will use the rule $x^{a}* x^{b} =x^{a+b}$ to factorise the exponents:
$12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)$
The greatest common factor is thus $12xy^{2}$

3 0

3 years ago

Find the composition of

Rus_ich [418]

Answer:

[(x + 6), (y + 1)]

Step-by-step explanation:

Vertices of the quadrilateral ABCD are,

A → (-5, 2)

B → (-3, 4)

C → (-2, 4)

D → (-1, 2)

By reflecting the given quadrilateral ABCD across x-axis to form the image quadrilateral A'B'C'D',

Rule for the reflection of a point across x-axis is,

(x, y) → (x , -y)

Coordinates of the image point A' will be,

A(-5, 2) → A'(-5, -2)

From the picture attached, point E is obtained by translation of point A'.

Rule for the translation of a point by h units right and k units up,

A'(x+h, y+k) → E(x', y')

By this rule,

A'(-5 + h, -2 + k) → E(1, -1)

By comparing coordinates of A' and E,

-5 + h = 1

h = 6

-2 + k = -1

k = 1

That means

Rule for the translation will be,

[(x + 6), (y + 1)]

8 0

2 years ago

Average speed of driving 330 miles in 5 and a half hours

ehidna [41]

The Average speed would be 60 miles per hour.

- Divide 330 by 5.5 hours

8 0

3 years ago