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3 years ago

Determine whether each equation has one solution, no solution, or infinite solutions.

Mathematics

1 answer:

Sati [7]3 years ago

6 0

Answer:

a. has one solution

b. infinite solution

Step-by-step explanation:

2(x - 1) + 6 = 4x - 22

2x - 2 + 6 = 4x - 22

2x - 4x = 2 - 6 - 22

-2x = -26

x = 26/2

x = 13

6(2x + 1) – 2 = 12x + 4

12x + 6 - 2 = 12x + 4

12x - 12x = 4 + 2 - 6

0 = 0

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Will give brainliestt

Natasha_Volkova [10]

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Step-by-step explanation:

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3 years ago

at noon joice drove to the lake at 30 mi. per hour, but she made a long walk back at 4 mi. per hour. how long did she walk ifshe

Kryger [21]

Distance = 30 mph * time
distance = 4 mph * time

distance = 30 mph * t
distance = 4 mph * (17-t)

Since distance is equal then
30 mph * t = 4 mph * (17-t)

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Double Check
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3 years ago

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 $s = \frac{\sigma}{\sqrt{n}}$ .