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Serga [27]
3 years ago
5

Pliz help solve this what's 3+4×2.00×29 Here the other question 5n+1 find the sequence

Mathematics
2 answers:
givi [52]3 years ago
6 0

Answer:

Step-by-step explanation:

1) 3 + 4*2*29 = 3 + 232  = 235  

First we have to do multiplication and then we have to add

2) 5n + 1

Term 1 ; n = 1  ;  5n + 1 = 5*1 + 1 = 5 + 1 = 6

Term 2 ; n = 2 ; 5n + 1 = 5*2 + 1 = 10 + 1 = 11

Term 3 ; n = 3 ; 5n+ 1 = 5*3 + 1 = 15 + 1 = 16

Term 4; n = 4  ; 5n + 1 = 5*4 + 1 = 20 + 1 = 21

The sequence : 6 , 11, 16 , 21 ..........

Stels [109]3 years ago
3 0
3+4x2.00x29 is 235.
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Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1
svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal 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.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that \mu = 8, \sigma = 1.1

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

n = 5 means that s = \frac{1.1}{\sqrt{5}} = 0.4919

This probability is the pvalue of Z when X = 9.3. So

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

By the Central Limit Theorem

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

Z = \frac{9.3 - 8}{0.4919}

Z = 2.64

Z = 2.64 has a pvalue of 0.9959

0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?

n = 6 means that s = \frac{1.1}{\sqrt{6}} = 0.4491

This probability is 1 subtracted by the pvalue of Z when X = 9. So

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

Z = \frac{9 - 8}{0.4491}

Z = 2.23

Z = 2.23 has a pvalue of 0.9871

1 - 0.9871 = 0.0129

0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

8 0
3 years ago
HEY I NEED HELP FOR DIS QUESTION. I WILL MARK BRAINLIEST FOR THE BEST ANSWER. PLEASE SHOW ALL WORK AND FULL SOLUTIONS. The volum
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9514 1404 393

Answer:

  12.0 cm

Step-by-step explanation:

Use the formula for the volume of a cylinder. Fill in the given values and solve for the height.

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  h = 1846/(49π) cm ≈ 12.0 cm . . . . . . divide by the coefficient of h

The height of the can is about 12.0 cm.

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e^{-x^2} has no antiderivative in terms of elementary functions (polynomials, exponentials, logarithms, trigonometric functions, etc), but there is a special function defined to fit that role called the error function, \mathrm{erf}(x), where

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