6x^2 - 37x + 13 = 6

1 answer:

6 0

You can easily apply it in a fx-570es Calculator and you’ll get

X=5.971
X=0.195

it’s hard to factor it because there’s no multiples of 42 that when you add them gives you 37 so that’s why the solutions are decimals

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Lena [83]

Answer:

Step-by-step explanation:

it is not strait

Hope this works!

7 0

3 years ago

5/8 + 4/5 = 9/8 1 17/40 57/40 9/13

valina [46]

Answer:

57/40

Step-by-step explanation:

Find common denominators, note that what you do to the denominator, you must do to the numerator:

$\frac{numerator}{denominator}$

First, multiply 5/8 with 5 to both the numerator and denominator:

(5/8) x (5/5) = 25/40

Next, multiply 4/5 with 8 to both the numerator and denominator:

(4/5) x (8/8) = 32/40

Combine the numerators:

(25 + 32)/40 = (57)/40

57/40 is your answer.

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2 years ago

Help!! I need answer ASAP!!!

Zolol [24]

Answer:

11 cm. if your first base is 11cm, and your second base is 13, then you need height. A=(a+b)/2 h=(13+11)/2 × 11= 132

5 0

3 years ago

A brewery has a beer dispensing machine that dispenses beer into the company's 12 ounce bottles. The distribution for the amount

-Dominant- [34]

Answer:

The company should use a mean of 12.37 ounces.

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 distribution for the amount of beer dispensed by the machine follows a normal distribution with a standard deviation of 0.17 ounce.

This means that $\sigma = 0.17$

The company can control the mean amount of beer dispensed by the machine. What value of the mean should the company use if it wants to guarantee that 98.5% of the bottles contain at least 12 ounces (the amount on the label)?

This is $\mu$ , considering that when $X = 12$ , Z has a p-value of $1 - 0.985 = 0.015$ , so when $X = 12, Z = -2.17$ .