Make x the main subject<br> 6x+a=5(x+t)

Kayla is on level 12 of a new game. Everyday she completes 6 new levels. Maddy is on level 28 of the same game. Everyday she com

ankoles [38]

Answer:

Kayla and Maddy will be in the same level in 4 days.

Step-by-step explanation:

Kayla: y = 6x + 12

Maddy: y = 2x + 28

6x + 12 = 2x + 28

Subtract 2x from both sides;

4x + 12 = 28

Subtract 12 from both sides;

4x = 16

Divide both sides by 4;

x = 4

3 0

2 years ago

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Find p and q for which the linear eqn has infinite solutions

jekas [21]

Answer:

p= 2.5

q= 7

Step-by-step explanation:

The lines should overlap to have infinite solutions, slopes should be same and y-intercepts should be same.

Equations in slope- intercept form:

6x-(2p-3)y-2q-3=0 ⇒ (2p-3)y= 6x -2q-3 ⇒ y= 6/(2p-3)x -(2q+3)/(2p-3)

12x-( 2p-1)y-5q+1=0 ⇒ (2p-1)y= 12x - 5q+1 ⇒ y=12/(2p-1)x - (5q-1)/(2p-1)

Slopes equal:

6/(2p-3)= 12/(2p-1)

6(2p-1)= 12(2p-3)

12p- 6= 24p - 36

12p= 30

p= 30/12

p= 2.5

y-intercepts equal:

(2q+3)/(2p-3)= (5q-1)/(2p-1)

(2q+3)/(2*2.5-3)= (5q-1)/(2*2.5-1)

(2q+3)/2= (5q-1)/4

4(2q+3)= 2(5q-1)

8q+12= 10q- 2

2q= 14

q= 7

7 0

2 years ago

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

Answer fast for pts

sdas [7]

Answer:

Since she arrived at 8:30 its not accurate because her records started at the time she arrived so she didnt get the times of the students that arrived before her

Step-by-step explanation:

Hope this helps:))

4 0

2 years ago

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9. The sum of 10 and 7 times a number is 59. What is the<br> number? (Use n as the variable.)

PSYCHO15rus [73]

10+7n=59
7n=59-10
7n=49
n=49/7
n=7

4 0

2 years ago

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Make x the main subject 6x+a=5(x+t)