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Vitek1552 [10]
3 years ago
15

The first step to solve the equation StartFraction x over 4 EndFraction minus three-fourths = 16 is shown below.

Mathematics
1 answer:
irina1246 [14]3 years ago
4 0
Ok so the first step to solve it is subtraction then it’s multiplication then additions
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A computer system password requires five characters, all lowercase letters (remember there are 26 letters), and all the letters
Goshia [24]
26 divided by 5 = 5.2
3 0
3 years ago
Factor completely 2x5 + 10x4 - 22x3.
PIT_PIT [208]

Answer:  2x³(x² + 5x - 11)

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

2x⁵ + 10x⁴ - 22x³

Factor out the GCF (2x³)

2x³(x² + 5x - 11)

Since there are no values whose product is -11 and sum is +5, this cannot be factored further.

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3 years ago
Can someone tell me what the blue lines mean?
Radda [10]

Answer:

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7 0
3 years ago
Read 2 more answers
The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

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

By the Central limit theorem

Z = \frac{X - \mu}{s}

Z = \frac{70.5 - 69}{1.97565}

Z = 0.76

Z = 0.76 has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0
3 years ago
Can some on e please help me understand this
umka21 [38]
It’s P.E.M.D.A.S

P= parenthesis ()
E= exponent
M= multiplication or Multiply
D= Division or divided
A= Addition
S= Subtraction
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


4 0
3 years ago
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