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zhannawk [14.2K]
3 years ago
6

Need Help solving this , pic of question

Mathematics
1 answer:
Mama L [17]3 years ago
7 0
<h3>Answer:</h3>

Length: 12 yd

Width: 5.5 yd

<h3>Explanation:</h3>

Let x represent the width of the rectangle in yards. Then 2x+1 can represent the length (double the width plus 1 yard). The area is the product of length and width so we have the relation ...

... x(2x+1) = 66

... 2x² +x -66 = 0

This can be solved using any of several methods. One of them is to factor the equation. To do that, we can look for factors of 2·66 = 132 that differ by 1. Since 132 = 11·12, those are the numbers we're looking for. Then ...

... 2x² +12x -11x -66 = 0 . . . . use our found numbers to rewrite the x term

... 2x(x +6) -11(x +6) = 0 . . . . .factor by grouping

... (2x -11)(x +6) = 0

... x = 11/2 or -6 . . . . . values that make the factors zero. Only the positive value is a useful solution.

... The width of the rectangle is 5.5 yards. Its length is 2·5.5+1 = 12 yards,

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Step-by-step explanation:

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A jet ski rental charges a flat rate of $75, plus an additional $60 per hour. You do not want to spend more than $200. How many
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22.5

Step-by-step explanation:

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3 years ago
BRAINLIESTT ASAP! PLEASE HELP ME :)<br><br> Solve for x in the figure below. Show your work.
nikdorinn [45]

Answer:

x = 19°

Step-by-step explanation:

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hence

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7 0
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Read 2 more answers
Select the most accurate statement regarding the normal distribution. Group of answer choices It is always the appropriate distr
PSYCHO15rus [73]

Answer:

There is a 95% chance that values will be within ±2 standard deviations of the mean.

Step-by-step explanation:

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this question:

According to the empirical rule, 95% of the measures are within 2 standard deviations of the mean. So the correct statement is:

There is a 95% chance that values will be within ±2 standard deviations of the mean.

4 0
3 years ago
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