Mia's cat weighs 17.2 pounds. Her dog weighs 3.1 times as much as her cat. How much does Mia's dog weigh?

Identify the zeros for the function and sketch a graph. y=x(x-2)(x+3)

damaskus [11]

The product is equal to zero if one of the factors is equal to zero.

Therefore:

$y=x(x-2)(x+3)\\\\x(x-2)(x+3)=0\iff x=0\ \vee\ x-2=0\ \vee\ x+3=0\\\\\boxed{x=0\ \vee\ x=2\ \vee\ x=-3}$

5 0

3 years ago

Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial t

SCORPION-xisa [38]

Answer:

(12.141, 14.059)

Step-by-step explanation:

Explanation is provided in the attached document.

Download docx

5 0

3 years ago

A bin contains 25 light bulbs, 5 of which are in good condition and will function for at least 30 days, 10 of which are partiall

Ira Lisetskai [31]

Answer:

The probability that it will still be working after one week is $\frac{1}{5}$

Step-by-step explanation:

Given :

Total number of bulbs = 25

Number of bulbs which are good condition and will function for at least 30 days = 5

Number of bulbs which are partially defective and will fail in their second day of use = 10

Number of bulbs which are totally defective and will not light up = 10

To find : What is the probability that it will still be working after one week?

Solution :

First condition is a randomly chosen bulb initially lights,

i.e. Either it is in good condition and partially defective.

Second condition is it will still be working after one week,

i.e. Bulbs which are good condition and will function for at least 30 days

So, favorable outcome is 5

The probability that it will still be working after one week is given by,

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

$\text{Probability}=\frac{5}{25}$

$\text{Probability}=\frac{1}{5}$

5 0

3 years ago

Mark earns $300 per month plus an additional 2% on all of his sales. Last month he sold $59,000. What was Mark’s income for the

Lina20 [59]

Answer:

59,000 * .02= $1,180 commission plus $300 = $1480

3 0

3 years ago

The average expenditure on Valentine's Day was expected to be$100.89 (USA Today, February 13, 2006). Do male and femaleconsumers

stealth61 [152]

Answer:

(a) $62.16

(b) Male: $15.00

Female: $10.06

(c) Confidence Interval for male expenditure is ($106.40, $136.40)

Confidence interval for female expenditure is ($49.18, $69.30)

Step-by-step explanation:

(a) Male expenditure

Sample mean = $135.67, sd=$35, n=40, Z=2.576

Population mean = sample mean - (Z×sd)/√n = 135.67 - (2.576×35)/√40 = 135.67 - 14.27 = $121.40

Female expenditure

Sample mean= $68.64, sd=$20, n=30, Z=2.576

Population mean = 68.64 - (2.576×20)/√30 = 68.64 - 9.40 = $59.24

$121.40 - $59.24 = $62.16

(b) Male: Error margin = (t-value × sd)/√n

Degree of freedom = n-1 = 40-1= 39. t-value corresponding to 39 degrees of freedom and 99% confidence level is 2.708

Error margin = (2.708×35)/√40 = 94.78/6.32 = $15.00

Female

Degrees of freedom = n-1 = 30-1 = 29. t-value is 2.756

Error margin = (2.756×20)/√30 = 55.12/5.48 = $10.06

(c) Male

Confidence Interval (CI) = (mean + or - error margin)

CI = 121.4 + 15.00 = $136.40

CI = 121.4 - 15.00 = $106.40

Confidence Interval is ($106.40, $136.40)

Female

CI = 59.24 + 10.06 = $69.30

CI = 59.24 - 10.06 = $49.18

Confidence Interval is ($49.18, $69.30)

7 0

4 years ago