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3 years ago

For what value of p will the pair of linear equations have infinitely many solutions (p-3)x + 3y = p , px+ py = 12

Mathematics

1 answer:

sleet_krkn [62]3 years ago

5 0

Answer: p = 6

Step-by-step explanation:

HI !

px + 3y - ( p-3)=0

a₁ = p , b₁ = 3 , c₁ = - (p-3)

=====================

12x + py - p=0

a₂ = 12 , b₂ = p , c₂ = -p

since , the equations have infinite solutions ,

a₁/a₂ = b₁/b₂ = c₁/c₂

p/12 = 3/p = p-3/p

--------------------------------------

p/12 = 3/p

cross multiply ,

p² = 36

p = √36

p = 6

for the value of p = 6 , the equations will have infinitely many solutions

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A book drive recorded its daily results on a graph. The first five days are represented as points (1,8), (2,13), (3,13), (4,15),

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The input of days results in the output of total collected books. On day 1, 8 books were collected. Both inputs for days 2 and 3 have the same output of 13, meaning that no books were collected for day 3. After 5 days, 21 books were collected. The most books were collected on day 5.

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3 years ago

Read 2 more answers

For the graph, what is a reasonable constraint so that the function is at least 200?

Anastasy [175]

Answer:

$0\leq x\leq 15$

Step-by-step explanation:

I think your question missed key information, allow me to add in and hope it will fit the orginal one

For the graph below, what should the domain be so that the function is at least 200? graph of y equals minus 2 times the square of x plus 30 times x plus 200

My answer:

Given the above information, we have:

$y=-2x^2+30x+200$

To make the function is at least 200, it means that:

$y=-2x^2+30x+200$ ≥ 200

<=> $-2x^2+30x$ ≥ 0

<=> x(-2x+30) ≥ 0

This is the product of two numbers hence would be positive only if either both are positive or both are negative

Case I: Both positive

x ≥ 0 and (-2x+30) ≥ 0

<=> 0 ≤ x ≤ 15

Case II: Both negative

Then we get

$x\leq 0 and -2x+30\leq 0\\\\x\leq 0 and x\geq 15$

This is inconsistent as a value cannot be less than 0 and greater than 15

=> our correct answer is $0\leq x\leq 15$

Hope it will find you well.

3 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05