All categories

3 years ago

I need help with this question please answer asap

Mathematics

1 answer:

saveliy_v [14]3 years ago

8 0

Answer:

9 months

Step-by-step explanation:

8 sculptures in 6 months is 1 sculpture in 0.75 months.

8/8=1

6/8=0.75

1×12=12

0.75×12=9

You might be interested in

How many equivalence relations are there on the set 1, 2, 3]?

Alex787 [66]

Answer:

We need to find how many number of equivalence relations are on the set {1,2,3}

A relation is an equivalence relation if it is reflexive, transitive and symmetric.

equivalence relation R on {1,2,3}

1.For reflexive, it must contain (1,1),(2,2),(3,3)

2.For transitive, it must satisfy: if (x,y)∈R then (y,x)∈R

3. For symmetric, it must satisfy: if (x,y)∈R,(y,z)∈R then (x,z)∈R

Since (1,1),(2,2),(3,3) must be there is R, (1,2),(2,1),(2,3),(3,2),(1,3),(3,1). By symmetry,

we just need to count the number of ways in which we can use the pairs (1,2),(2,3),(1,3) to construct equivalence relations.

This is because if (1,2) is in the relation then (2,1) must be there in the relation.

the relation will be an equivalence relation if we use none of these pairs (1,2),(2,3),(1,3) . There is only one such relation: {(1,1),(2,2),(3,3)}

we can have three possible equivalence relations:

{(1,1),(2,2),(3,3),(1,2),(2,1)}

{(1,1),(2,2),(3,3),(1,3),(3,1)}

{(1,1),(2,2),(3,3),(2,3),(3,2)}

6 0

4 years ago

ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply to them for insurance fall into these th

Misha Larkins [42]

Answer:

a. $P(E_1/A)=0.0789$

b. $P(E_2/A)=0.395$ \

c. $P(E_3/A)=0.526$

Step-by-step explanation:

Let $E_1,E_2,E_3$ are the events that denotes the good drive, medium drive and poor risk driver.

$P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20$

Let A be the event that denotes an accident.

$P(A/E_1)=0.01$

$P(A/E_2=0.03$

$P(A/E_3)=0.10$

The company sells Mr. Brophyan insurance policy and he has an accident.

a.We have to find the probability Mr.Brophy is a good driver

Bayes theorem, $P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}$

We have to find $P(E_1/A)$

Using the Bayes theorem

$P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$

Substitute the values then we get

$P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}$

$P(E_1/A)=0.0789$

b.We have to find the probability Mr.Brophy is a medium driver

$P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395$

c.We have to find the probability Mr.Brophy is a poor driver

$P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526$

7 0

4 years ago

The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month constantly for its first year.

iris [78.8K]

Answer is D if the function W is graphed find and interpret the slope of the function

6 0

3 years ago

Craig went bowling with $25 to spend he rented shoes for $5.25 and paid $4dollars for each game what was the greatest number of

BlackZzzverrR [31]

He would play 4 games.

3 0

3 years ago

Read 2 more answers

out of 45 times at bat, Raul got 19 hits.Find Rauls batting average as a decimal rounded tothe nearst thousandth.

kati45 [8]

19/45=0.422222222........
Therefore rounded to the nearest thousandth his batting average is 0.422

3 0

3 years ago