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diamong [38]
3 years ago
7

HELP PLEASE ILL MARK BRAINLIEST

Mathematics
2 answers:
choli [55]3 years ago
4 0

Answer:

2.3

Step-by-step explanation:

did the math

dezoksy [38]3 years ago
3 0
So, basically the x is 2.3 !
You might be interested in
Which function is increasingnon the interval (negative infinity, positive infinity)
BlackZzzverrR [31]

Answer:

h(x) = 2^(x) - 1.

Step-by-step explanation:

Let's look at each equation:

f(x) = -3x +7, Well as x increases, since it's multiplication, there are "going to be more" -3's, so it's going to be decreasing.

g(x) = -4(2^x). While 2^x is increasing, because "there are going to be more 2's multiplied by each other" as x increases, it's being multiplied by a negative number, so it's actually going to be decrasing

h(x) = 2^(x) - 1. Here's it's going to be increases as x goes towards infinity because "there are going to be more 2's multiplied by each other", and there isn't any negative sign, while there is a negative 1, it's constant, so the overall value will be increasing

3 0
1 year ago
125 a la octava potencia? <br>please lo necesito ​
telo118 [61]
No se si esté correcto, buena suerte!:)

8 0
3 years ago
I need help with math equation: 5 1/4 a = 7/8<br> a =
kari74 [83]
Answer: a = 1/6
Explanation: Convert the mixed number into an improper fractions. Which is 21/4.
Then you can do the butterfly method.
6 0
2 years ago
A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the
Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 3.6, \sigma = 0.4

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

X = 3.6

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

Z = \frac{3.6 - 3.6}{0.4}

Z = 0

Z = 0 has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

Z = \frac{5 - 3.6}{0.4}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998

X = 4.2

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

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

Z = \frac{5 - 3.6}{0.4}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998

X = 3

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

Z = \frac{3 - 3.6}{0.4}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

1.75 = \frac{X - 3.6}{0.4}

X - 3.6 = 0.4*1.75

X = 4.3

They last at least 4.3 minutes

7 0
3 years ago
HELP ASAP PLEASE!!! This is a written response
o-na [289]

Answer:

Exponential decay.

Step-by-step explanation:

The magnitude of the base (also known as the "common factor") tells us whether a given exponential equation represents growth or decay.

Standard equation for expo. growth:  y = 3(b)^x, where b is the base.

If 0<b<1, we have expo. decay; if 1<b, we have expo. growth.

Thus, y = 3(0.5)^x represents expo. decay.

8 0
3 years ago
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