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3 years ago

The prime factorization of 16x5

Mathematics

1 answer:

RoseWind [281]3 years ago

6 0

Answer:

well first 16x5=80

2 * 2 * 2 * 2 * 5

Step-by-step explanation:

Hope that helps! :)

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A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu

telo118 [61]

Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

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 samples 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 110, p = 0.2$

$\mu = E(X) = np = 110*0.2 = 22$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952$

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

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

$Z = \frac{25 - 22}{4.1952}$

$Z = 0.715$

$Z = 0.715$ has a pvalue of 0.7626.

X = 17

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

$Z = \frac{17 - 22}{4.1952}$

$Z = -1.19$

$Z = -1.19$ has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

4 0

3 years ago

Calculate the sum of the first 30 terms in the sequence that begins with 70,65,60

Afina-wow [57]

$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\ddddddddddddddddddd \displaystyle \Large \boldsymbol{} a_1=70 \ ; \ d=-5 \\\\S_n=\frac{a_1+a_n}{2} \cdot n \ ; \ a_n=a_1+(n-1)d \Longrightarrow \\\\\\ S_{30}=\frac{70+70-29\cdot5}{2} \cdot 30 =-75 \\\\\\Answer: S_{30}=-75$

4 0

3 years ago

Which expression is equivalent to -1.3 - (-1.9)

Verdich [7]

Answer:

-8-(-1)-5

Step-by-step explanation:

(Hope this helps can I pls have brainlist (crown) ☺️)

5 0

3 years ago

Write (-5+8i)+(-5-8i) as a complex number in standard form

gizmo_the_mogwai [7]

$(-5+8i)+(-5-8i) =
-5+8i-5-8i=-10$

5 0

3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

3 years ago