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oksian1 [2.3K]
3 years ago
14

9/10 divided by (-3/5)

Mathematics
2 answers:
storchak [24]3 years ago
7 0

Answer: -1 and 1/2

Step-by-step explanation: It's important to understand that dividing by a fraction means the same thing as multiplying by the reciprocal of that fraction. In other words, we can simply change the division symbol to multiplication and flip the second fraction.

So we can rewrite this problem as \frac{9}{10} x \frac{-5}{3}.

Before multiplying two fractions together, you should always try to cross cancel and notice that here we can cross cancel the 9 and 3 to 1 and 3. Also notice that we are multiplying a positive times a negative which is a negative so we know that our answer is negative.

Now we can multiply across the numerators and across the denominators to get 3 times -5 or -15 over 10 times 1 or 10.

So our answer is -15/10.

Notice that our fraction is not in lowest terms so we could have cross canceled again. That's no problem though because we can still simplify. If we divide both top and bottom of our fraction by 5, we get -3/2.

Notice that we have an improper fraction so we can write it as a mixed number by dividing the denominator into the numerator and we get -1 with a remainder of 1 so we can write this as -1 1/2. We keep the denominator the same.

So 9/10 divided by -3/5 is -1 and 1/2.

Alex787 [66]3 years ago
3 0
The exact form is -3/2
The decimal form is -1.5
The mixed number form is -1 1/2
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Let u,v,wu,v,w be three linearly independent vectors in R7R7. Determine a value of kk, k=k= , so that the set S={u−3v,v−2w,w−ku}
11Alexandr11 [23.1K]

Answer:

1/6

Step-by-step explanation:

For a set of vectors a, b and c to be linearly dependent, the linear combination of the set of vectors must be zero.

C1a + C2b + C3c = 0

From the question, we are given the set S={u−3v,v−2w,w−ku}, the corresponding vectors are a(0, -3, 1), b(-2,1,0) and c(1,0,-k)

The values in parenthesis are the ccoefficients of w, v and u respectively.

On writing this vector as a linear combination, we will have;

C1(0, -3, 1) + C2(-2,1,0) + C3(1,0,-k) = (0,0,0)

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-3C1+C2 = 0 ........... 2

C1-kC3 = 0 ….......... 3

From equation 2, 3C1 = C2

Substituting into 1, -2(3C1)+C3 = 0

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C1-k(6C1) = 0

C1 = k6C1

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Hence the value of k for the set of vectors to be linearly dependent is 1/6

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True

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However, increases in N cause both to decrease, since sampling error is reduced.

A small sample size might lead to frequent Type II errors, i.e. it could be that your (alternative) hypotheses are right, but because your sample is so small, you fail to reject the null even though you should.

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