What is the average of 66 and 149.

Mathematics

2 answers:

olganol [36]3 years ago

7 0

157.5 is the correct answer add 66 and 149 you get 315 divide 315 by 2 and you get 157.5

Minchanka [31]3 years ago

5 0

66 + 149 = 215
215/2 = 107.5
107.5 is the average.

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agasfer [191]

We use the Work formula to solve for the unknown in the problem which is W = F x d. First, we solve for the Net Force acting on the car. The Net Force is the summation of all forces acting on the object. For this case, we assume that Friction Force is negligible thus the Net Force is equal to:

F = mgsinα in terms of SI units and in terms of english units we have F = m(g/g₀)(sin α) where g₀ is the proportionality factor, 32.174 ft lb-m / lb-f s²

F = 2500 (32.174/32.174) (sin 12°) = 519.78 lb

W = Fd = 519.78 lb (400 ft) = 207912 ft - lb or 20800 ft-lb

5 0

3 years ago

The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resista

baherus [9]

Answer:

The mean is 9.65 ohms and the standard deviation is 0.2742 ohms.

Step-by-step explanation:

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.

10% of all resistors having a resistance exceeding 10.634 ohms

This means that when X = 10.634, Z has a pvalue of 1-0.1 = 0.9. So when X = 10.634, Z = 1.28.

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

$1.28 = \frac{10.634 - \mu}{\sigma}$