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Nataliya [291]
3 years ago
5

How many times does X squared minus 4X -12 cross the X axis

Mathematics
1 answer:
kolbaska11 [484]3 years ago
4 0

We can solve this problem using discriminant.

x^2-4x-12's discriminant is

(-4)^2-4*-12=16+48 which is clearly larger than 0

This means that it crosses over the axes 2 times.

In case you don't know what discriminant is, its in equation ax^2+bx+c

the discriminant is b^2-4ac.

If its positive it has 2 crosses with x axis, if negative then 0 crosses, if 0 then 1 cross.

Hope this helped at least a little bit :D

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Which expression is equivalent to 8a +14
andrey2020 [161]

Answer:

2(4a + 7)

Step-by-step explanation:

2 to the 4 and 7

This would give you 8a + 14 because 2 times 4 is 8,  and 2 times 7 is 14.

<em><u>Hope this helps.</u></em>

8 0
3 years ago
Read 2 more answers
According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax for
Wewaii [24]

Answer:

78.52% probability that the sample mean is greater than 320 minutes

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 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.

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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 330, \sigma = 80, n = 40, s = \frac{80}{\sqrt{40}} = 12.65

What is the likelihood the sample mean is greater than 320 minutes?

This is 1 subtracted by the pvalue of Z when X = 320. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{320 - 330}{12.65}

Z = -0.79

Z = -0.79 has a pvalue of 0.2148

1 - 0.2148 = 0.7852

78.52% probability that the sample mean is greater than 320 minutes

8 0
3 years ago
26.8 +0.5w = 4
NemiM [27]
=45.6 rounded to the tenth
3 0
2 years ago
Read 2 more answers
The center of the circle is (-6,4). the circle passes through the point (-3,5). what is it’s radius?
alex41 [277]

Answer:

r = sqrt[ ( -3 - (-6))^2 + (5-4)^2 ] = sqrt(3^2 + 1^2)= sqrt(10)

Step-by-step explanation:

5 0
3 years ago
Please Help! I don't understand.
Andrei [34K]
C had the coordinate pair of (-2,2)
5 0
3 years ago
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