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3 years ago

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A farmer was counting his chickens and dogs. He counted a total of 60 legs. If the amount of chickens he has is four times the a

mount of dogs he has, how many chickens does he have?

1 answer:

Kay [80]3 years ago

7 0

Answer:

he has 20 chickens

Step-by-step explanation:

i was testing with it. 20 dog legs divided by 4 (since dogs have 4 legs) equals 5. 5 times 4 (since 4 times the amount of dogs) is 20. then to check, 20 times 2 (since chickens have 2 feet) is 40. 20 dog legs plus 40 chicken legs. makes sense. again, 20 chickens.

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7n-(4n-3) combining like terms with negative coefficient and distribution

sladkih [1.3K]

$\huge\text{Hey there!}$

$\huge\textbf{Equation/Question:}$

$\mathbf{7n - (4n - 3)}$

$\huge\textbf{Convert equation/question to:}$

$\mathbf{= 7n - 1(4n - 3)}$

$\huge\textbf{Solving the equation/question:}$

$\mathbf{= 7n - 1(4n) - 1(-3)}$

$\mathbf{= 7n - 4n + 3}$

$\huge\textbf{Combine the like terms:}$

$\mathbf{= (7n - 4n) + 3}$

$\mathbf{= 7n - 4n + 3}$

$\mathbf{= 3n + 3}$

$\huge\textbf{Therefore, your answer should be:}$

$\huge\boxed{\frak{3n + 3}}\huge\checkmark$

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<h3>~ $\frak{Amphitrite1040:)}$ </h3>

6 0

1 year ago

Factor the polynomial 2x^3+4x^2+8x

Sati [7]

So you want to take out the GCF first.
The GCF is 2x
2x(x^2+2x+4)
This is your final answer because there is not other way to factor.

3 0

2 years ago

Read 2 more answers

Rita, Abdul, and Deon have a total of $ 128 in their wallets. Abdul has $8 less than Rita. Deon has 3 times what Abdul has

Charra [1.4K]

Answer:

They have all together 1,208$

Step-by-step explanation:

Do subtraction first, 128- 8= 120 so Abdul as 120$

Deon is multiplication, 8x120=960$

Rita has 128$

3 0

2 years ago

According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e

slamgirl [31]

Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

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.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that $p = 0.09$

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

So

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225$

$P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827$

$P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082$

$P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982$

Finally

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180$

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

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2 years ago

I need the answer ...

solong [7]

Ooh so the directions tell us what we need to find the entire line, which is made up of FH. so just add GH (15) and FG (6) together to find FH. so that is 21. FH=21.

3 0

2 years ago