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3 years ago

Multiplying binomials Determine the product (x-3) (x-8)

Mathematics

2 answers:

Setler79 [48]3 years ago

7 0

Answer:

x^2 - 11x + 24

Step-by-step explanation:

(x-3)(x-8)

x*x = x^2 x*-8 = -8x

-3*x = -3x -3*-8 = 24

x^2 - 8x - 3x +24

x^2 -11x +24

MrMuchimi3 years ago

4 0

X^2-24 is the answer

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at a movie theater adult tickets cost $8.00 and student tickets cost $5.00. If a big family bought 9 tickets and spent $48.00, h

Anastasy [175]

Answer:

8 student tickets and 1 adult ticket

Step-by-step explanation:

5 times 8 is 40

40 plus 8 is 48

which gives you $48.00

5 0

3 years ago

The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand

Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

$P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})$

$P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})$

$P(X > 1560) = P(Z > \dfrac{110}{220})$

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean $\mu_x$ = np = 52 × 0.3085 = 16.042

The standard deviation = $\sqrt {n \time p (1-p)}$

The standard deviation = $\sqrt {52 \times 0.3085 (1-0.3085)}$

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

$Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})$

$Pr ( Y > 20) = P(Z >1 .338)$

From z tables

P(Y > 20) $\simeq$ 0.0903

7 0

3 years ago

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the propo

never [62]

Answer:

A sample of 997 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that $\pi = 0.08$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a p-value of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}$

$0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}$

$\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2$

$n = 996.3$

Rounding up:

A sample of 997 is needed.

3 0

3 years ago

1/2=4p what does p equal

disa [49]

P = what times 4 equals 1/2 = 1/8

3 0

3 years ago

A scale of 0.5 inches to 2 feet was used for a model of the goal posts on a football field. If the model was 1 1/2 in high and 5

Paraphin [41]

The actual height is: 1 1/2 * 2 = 3 feet
The actual length is: 5 1/8 * 2 = 10.25 feet

hope this helps you.

6 0

3 years ago