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4 years ago

5

A tiny area on a computer chip measures 2.3 mm by 1.7 mm. About how many times longer is the longer dimension than the shorter d

imension?

2 answers:

SVEN [57.7K]4 years ago

8 0

2.3=1.7x
x is how many times longer the longer side is.
just divide 2.3 by 1.7
it is 1.35294...
round to nearest 10th im assuming, so
it is 1.4 or round however you are told to round.

olga2289 [7]4 years ago

5 0

<h2>Answer:</h2>

Hence, the longer dimension is approx 1.4 times longer than the shorter dimension.

<h2>Step-by-step explanation:</h2>

The tiny area on a computer chip measures 2.3 mm by 1.7 mm.

This means that the shorter dimensions measure: 1.7 mm

and the longer dimension measures: 2.3 mm

Now, let the longer dimension be x times the shorter dimension.

i.e.

$2.3=1.7x\\\\i.e.\\\\x=\dfrac{2.3}{1.7}\\\\i.e.\\\\x=1.353$

which is approximately equal to 1.4 times.

Hence, the answer is: approx 1.4 times.

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I need this done in 10mins

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3 years ago

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3 years ago

62% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

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Answer:

a) 0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

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Step-by-step explanation:

To solve this question, we use the binomial probability distribution, and also it's approximation to the normal distribution.

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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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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This means that $p = 0.62$

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This means that $n = 48$

Mean and standard deviation, for the approximation:

$\mu = E(x) = np = 48*0.62 = 29.76$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{48*0.62*0.38} = 3.36$

a. Exactly 30 of them are spayed or neutered.

This is P(X = 30), which is not necessary the use of the approximation.

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

$P(X = 30) = C_{48,30}.(0.62)^{30}.(0.38)^{18} = 0.1180$

0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b. At most 33 of them are spayed or neutered.

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$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{33.5 - 29.76}{3.36}$

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$Z = 1.11$ has a pvalue of 0.8665

0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c. At least 31 of them are spayed or neutered.

Using continuity correction, this is $P(X \geq 31 - 0.5) = P(X \geq 30.5)$ , which is 1 subtracted by the pvalue of Z when X = 30.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30.5 - 29.76}{3.36}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

1 - 0.5871 = 0.4129

0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d. Between 24 and 30 (including 24 and 30) of them are spayed or neutered.

This is, using continuity correction, $P(24 - 0.5 \leq X \leq 30 + 0.5) = P(23.5 \leq X \leq 30.5)$ , which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 23.5.

X = 30.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30.5 - 29.76}{3.36}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

X = 23.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{23.5 - 29.76}{3.36}$

$Z = -1.86$

$Z = -1.86$ has a pvalue of 0.0314

0.5871 - 0.0314 = 0.5557

0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

8 0

3 years ago

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Jet001 [13]

Answer:

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2 years ago