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3 years ago

Which values of a, b, and c represent the answer in simplest form? 7/9 / 4/9

Mathematics

2 answers:

satela [25.4K]3 years ago

8 0

1 3/4 the answer is

alina1380 [7]3 years ago

4 0

I got 1 1/4 hope that helps if u need me to tell u how I’ll show u

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PROBLEMAS QUE IMPLICAN UNA FUNCION DE 1ER GRADO

WINSTONCH [101]

anse is wter ixe 6 e8vd d Step-by-step explanation:

3 0

3 years ago

The student body President of a high school claims to know the names of at least 1000 of the 1800 students at the school. To tes

hoa [83]

Answer:

The 99% confidence interval for p in this case is (0.3317, 0.5883).

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Randomly selects 100 students from the school and asks the President to name each one. The President is able to correctly name 46 of the students.

This means that:

$n = 100, \pi = \frac{46}{100} = 0.46$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 - 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.3317$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 + 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.5883$

The 99% confidence interval for p in this case is (0.3317, 0.5883).

4 0

2 years ago

Brandon currently runs the mile in 6.8 minutes.

quester [9]

6 mins 38 seconds

.8 of 60 secs is 48 seconds so his time is 6 mins and 48 seconds, cutting off 10 seconds gives a new time of 6 mins 38 seconds

7 0

3 years ago

I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.

masya89 [10]

Answer:

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

4 0

2 years ago

Liza needs a total of 22.23 square feet of cloth to make a bag and a towel the bag requires 5.13 square feet of cloth the height

Troyanec [42]

Answer:

5.7 feet

Step-by-step explanation:

Total cloth area recquired to make a bag and a towel is 22.23 square feet

Given the area of the cloth recquired to make the bag is 5.13 square feet.

The area of cloth recquired to make the towel is = 22.23-5.13=17.1 square feet

given the height of the towel = 3 feet

let the length of the towel be x

We know that $area = length\times height$

area=3x

$3x=17.1\\x=5.7$

therefore the length of the towel is 5.7 feet

6 0

3 years ago