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3 years ago

First to answer gets brainlest

Mathematics

1 answer:

Bezzdna [24]3 years ago

7 0

Answer:

B) 279.24

Step-by-step explanation:

line them up the multiple then move the decimal

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WILL GIVE BRAINLIEST TO THE RIGHT ANSWER

Kryger [21]

It depends on the vehicle but normally four of five

8 0

3 years ago

Read 2 more answers

How to find the equation?

bixtya [17]

Answer:

y = 2x + 60

Step-by-step explanation:

Use the equation for a line: y = mx + b

Where

x and y are points on the graph

and m is the slope, and can be found by finding $\frac{rise}{run}$

so the slope is 10/5 or 2, since the graph goes up by 10 and over by 5.

so plug into the equation, and you get y = 2x + b

Next, pick a point on the graph, and plug in the x and y coordinates,

for example:

(x,y) = (5,70)

so, 70 = 2(5) + b

then solve for b

60 = b

Then rewrite the equation

y = 2x + 60

4 0

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