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2 years ago

Manda converted the following repeating decimal a fraction. Her work is shown below.

Mathematics

1 answer:

Butoxors [25]2 years ago

6 0

In analyzing Manda’s work, she made a mistake instep 3, she did not subtract one x from the left side.

<h3>What is the fraction about?</h3>

Fractions are known to be the parts of a complete or whole numbers of objects. It is known to have a numerator and a denominator and when solving for it, the rule of BODMAS must be obeyed.

In step 3 we see that (image attached) she did not subtract one x from the left side. The right thing to do is to use x to cancel out on the both side and that will make the answer to be correct but she did not.

So her error is seen in step 3 and as such, option C is correct.

See full question in the image attached.

Options are

Analyze Manda’s work. Is she correct? If not, what was her mistake?

A. Yes, she is correct.

B. In step 2, she needed to multiply both sides by 10.

C. In step 3, she did not subtract one x from the left side.

D. In step 4, she did not simplify her fraction.

Learn more about fraction from

brainly.com/question/78672

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Memory module consists of 9 chips. The device is designed with redundancy so that it works even if one of its chips is defective

soldier1979 [14.2K]

Answer:

a) $P[C]=p^n$

b) $P[M]=p^{8n}(9-8p^n)$

c) n=62

d) n=138

Step-by-step explanation:

Note: "Each chip contains n transistors"

a) A chip needs all n transistor working to function correctly. If p is the probability that a transistor is working ok, then:

$P[C]=p^n$

b) The memory module works with when even one of the chips is defective. It means it works either if 8 chips or 9 chips are ok. The probability of the chips failing is independent of each other.

We can calculate this as a binomial distribution problem, with n=9 and k≥8:

$P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)$

$P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9$

This equation was solved graphically and the result is that the maximum number of chips to have a reliability of the memory module equal or bigger than 0.9 is 62 transistors per chip. See picture attached.

d) If the memoty module tolerates 2 defective chips:

$P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2$

We again calculate numerically and graphically and determine that the maximum number of transistor per chip in this conditions is n=138. See graph attached.