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2 years ago

If 6 burgers cost $7.74 how many can you buy for $19.35

Mathematics

1 answer:

xeze [42]2 years ago

6 0

Answer:

15 burgers!

Step-by-step explanation:

The trick is to get rid of the decimal point from the number we are dividing by.

How? We can "shift the decimal point" out of the way by multiplying by 10, as many times as we need to.

But we must do the same thing to both numbers in the division.

Example: 15 divided by 0.2

Let us multiply the 0.2 by 10, which shifts the decimal point out of the way:

0.2 × 10 = 2

But we must also do it to the 15:

15 × 10 = 150

So 15 ÷ 0.2 has become 150 ÷ 2 (they are both 10 times larger):

150 ÷ 2 = 75

And so the answer is:

15 ÷ 0.2 = 75

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Help pls I have a test due tomorrow :(

Mademuasel [1]

Good luck on your test love ya also I have bad period cramps send chocolate please

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2 years ago

What is the goal of a proof by contradiction?

kifflom [539]

Answer:

B. The contradiction of the statement is proven false, so the original

statement is therefore proven true.

Step-by-step explanation:

We assume the opposite and then get to a point where we get a false statement. That proves that the the opposite of the opposite ( what we originally wanted to prove) is true.

6 0

2 years ago

Read 2 more answers

The ampitude/frequency/period of y=1/4cos1/2x is 4 pie

Fantom [35]

Amplitude: 1/4 (= |a|)
Frequency: 1/2 (= |b|)
Period: 4pi (= 2pi/|b|)

7 0

3 years ago

If 90 percent of automobiles in Orange County have both headlights working, what is the probability that in a sample of eight au

Marta_Voda [28]

Answer:

$P(X \geq 7) = P(X=7) +P(X=8)$

And we can find the individual probabilities using the probability mass function

$P(X=7)=(8C7)(0.9)^7 (1-0.9)^{8-7}=0.3826$

$P(X=8)=(8C8)(0.9)^8 (1-0.9)^{8-8}=0.4305$

And replacing we got:

$P(X \geq 7) = P(X=7) +P(X=8)=0.3826 +0.4305=0.8131$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest "number of automobiles with both headligths working", on this case we now that:

$X \sim Binom(n=8, p=0.9)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

And for this case we want to find this probability:

$P(X \geq 7) = P(X=7) +P(X=8)$

And we can find the individual probabilities using the probability mass function

$P(X=7)=(8C7)(0.9)^7 (1-0.9)^{8-7}=0.3826$

$P(X=8)=(8C8)(0.9)^8 (1-0.9)^{8-8}=0.4305$

And replacing we got:

$P(X \geq 7) = P(X=7) +P(X=8)=0.3826 +0.4305=0.8131$

7 0

2 years ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago