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3 years ago

A donkey weighs 570 pounds and an elephant weighs 5 tons. How much more does the elephant weigh?

2 answers:

7 0

A ton is equivalent to 1000 pounds so:
the elephant weighs 5000 pounds
5000-570= 4430
The elephants weighs 4430 pounds more.

kogti [31]3 years ago

5 0

The correct answer is:

5 tons = 10,000 lb

10,000 lb − 570 lb = 9,430 lb = 4 tons and 1,430 lb

This answer is if you want it in tons and pounds.

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I will mark as the brainiest if you get this correct<br> thx so much xxx

fgiga [73]

360 g plain flour
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4 0

3 years ago

You are planning to make an open rectangular box from aa 4141-in.-by-8181-in. piece of cardboard by cutting congruent squares

marissa [1.9K]

Answer:

13058.83 cubic inches

Step-by-step explanation:

Given that a rectangular box is having dimensions as 41x81 inches.

Let x be the side of square cut from all the four corners.

The open box made would have height as x and length 41-2x with width 81-2x

Volume =

$V(x) = x(41-2x)(81-2x)\\V(x) =x(3321-244x+4x^2)\\V(x) = 3321x-244x^2+4x^3\\V'(x) = 3321-488x+12x^2\\V"(x) = -488+24x\\$

Equate first derivative to 0

We get applicable root as x = 8.642

Max volume = 13058.83 cubic inches

3 0

3 years ago

Seventy-two percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of th

Anton [14]

Answer:

a) 0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) 0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) 0.064 = 6.4% probability that 7 of them are discovered.

Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

$P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312$

Probability of having an emergency locator and not being discovered:

20% of 28%. So

$P(A cap B) = 0.2*0.28 = 0.056$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105$

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

$P(A) = 1 - 0.5312 = 0.4688$

Probability of not having an emergency locator and being discovered:

34% of 72%. So

$P(A \cap B) = 0.34*0.72 = 0.2448$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522$

0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

p is the probability of being discovered with the emergency recorder:

0.5312 probability of having the emergency recorder.

Probability of having the emergency recorder and being located:

66% of 72%. So

$P(A \cap B) = 0.66*0.72 = 0.4752$

Probability of being discovered, given that it has the emergency recorder:

$p = P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4752}{0.5312} = 0.8946$

This question asks for P(X = 7) when $n = 10$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064$

0.064 = 6.4% probability that 7 of them are discovered.

8 0

3 years ago

If f(x)=3x-2 and g(x)=x^2+1, find f(g(-1)) and g(f(3))

julsineya [31]

Answer:

$f(g(-1))=4$

$g(f(3))=50$

Step-by-step explanation:

We know the definition of both functions: $f(x)=3x-2$ , and $g(x)=x^2+1$

A) In order to evaluate what $f(g(-1)) is$ , let's first investigate what g(-1) is using the definition for this function:

$g(x)=x^2+1\\g(-1)=(-1)^2+1\\g(-1)=1+1\\g(-1)=2$

Now let's find what f(2) is using f(x) definition: $f(x)=3x-2\\f(2)=3(2)-2\\f(2)=6-2\\f(2)=4$

B) In order to evaluate what $g(f(3)) is$ , let's first investigate what f(3) is using the definition for this function:

$f(x)=3x-2\\f(3)=3(3)-2\\f(3)=9-2\\f(3)=7$

Now let's find what g(7) is using the definition for this function:

$g(x)=x^2+1\\g(7)=(7)^2+1\\g(7)=49+1\\g(7)=50$

4 0

3 years ago

A code consists of two digit numbers chosen from 00-99 (i.E 00, 01, 02…..99) followed by two different letters of the alphabet.

Ede4ka [16]

Answer: 1/67600

Step-by-step explanation:

probability that the code is 43MZ:

Probability = (Total Required outcome / total possible outcomes)

Possible outcomes (digit) = (00,.............. ..., 99)

2.) possible outcomes (alphabet) = (A, B,............... .., Z)

P(43MZ) = P(43) × P(M) × P(Z)

P(43) = 1/100

P(M) = 1/26

P(Z) = 1/26

THEREFORE ;

P(43MZ) = (1/100) × (1/26) × (1/26) = 1/67600

4 0

4 years ago