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2 years ago

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1. Write an equation in slope-intercept form to represent the situation. Leslie joins a cooking club with a one time fee of $55.

She then pays $25 per month Your answer

1 answer:

mihalych1998 [28]2 years ago

7 0

Answer:

y = 25x + 55

Step-by-step explanation:

You might be interested in

(see image for question)

almond37 [142]

Answer:

Hans has done a mistake in the second step.

The correct answer is:

5x^2+13x-6=0

5x^2+15x-2x-6x=0 - 5. -6=-30 -30=15. -2

5x(1+3x)-2x(1+3x)=0

(5x-2x)(1+3x)=0

5-2x=0 1+3x=0

x=5/2 x=-1/3

4 0

2 years ago

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

$P(X\leq 1) = P(X=0) + P(X =1)$

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

$P(X\leq 1) = P(X=0) + P(X =1)$

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

6 0

2 years ago

How do you solve this?

solong [7]

It looks like your equations are

7M - 2t = -30

5t - 12M = 115

<u>Solving by substitution</u>

Solve either equation for one variable. For example,

7M - 2t = -30 ⇒ t = (7M + 30)/2

Substitute this into the other equation and solve for M.

5 × (7M + 30)/2 - 12M = 115

5 (7M + 30) - 24M = 230

35M + 150 - 24M = 230

11M = 80

M = 80/11

Now solve for t.

t = (7 × (80/11) + 30)/2

t = (560/11 + 30)/2

t = (890/11)/2

t = 445/11

<u>Solving by elimination</u>

Multiply both equations by an appropriate factor to make the coefficients of one of the variables sum to zero. For example,

7M - 2t = -30 ⇒ -10t + 35M = -150 … (multiply by 5)

5t - 12M = 115 ⇒ 10t - 24M = 230 … (multiply by 2)

Now combining the equations eliminates the t terms, and

(-10t + 35M) + (10t - 24M) = -150 + 230

11M = 80

M = 80/11

It follows that

7 × (80/11) - 2t = -30

560/11 - 2t = -30

2t = 890/11

t = 445/11

4 0

1 year ago

At a university, 60% of the 7,400 students are female. The student newspaper reports the results of a survey of a random sample

omeli [17]

Given Information:

Population mean = p = 60% = 0.60

Population size = N = 7400

Sample size = n = 50

Required Information:

Sample mean = μ = ?

standard deviation = σ = ?

Answer:

Sample mean = μ = 0.60

standard deviation = σ = 0.069

Step-by-step explanation:

We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10

np ≥ 10

50*0.60 ≥ 10

30 ≥ 10 (satisfied)

n(1 - p) ≥ 10

50(1 - 0.60) ≥ 10

50(0.40) ≥ 10

20 ≥ 10 (satisfied)

The mean of the sampling distribution will be same as population mean that is

Sample mean = p = μ = 0.60

The standard deviation for this sampling distribution is given by

$\sigma = \sqrt{\frac{p(1-p)}{n} }$

Where p is the population mean that is proportion of female students and n is the sample size.

$\sigma = \sqrt{\frac{0.60(1-0.60)}{50} }\\\\\sigma = \sqrt{\frac{0.60(0.40)}{50} }\\\\\sigma = \sqrt{\frac{0.24}{50} }\\\\\sigma = \sqrt{0.0048} }\\\\\sigma = 0.069$

Therefore, the standard deviation of the sampling distribution is 0.069.

4 0

2 years ago

Five times the sum of a number and 27 is greater than or equal to six times the sum of that num

stepladder [879]

Question:

Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26.

What is the solution set of this problem?

Answer:

$x \leq -21$

Step-by-step explanation:

Given

<em>Represent the number with x</em>

So:

$5 * (x + 27) \geq 6 * (x + 26)$

Required

Determine the solution set

$5 * (x + 27) \geq 6 * (x + 26)$

Open Both Brackets

$5 * x + 5 * 27 \geq 6 *x + 6 * 26$

$5x + 135 \geq 6x + 156$

Collect Like Terms

$5x - 6x \geq 156-135$

$- x \geq 21$

Multiply both sides by -1

$x \leq -21$

Hence, the solution set is $x \leq -21$

7 0

2 years ago