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padilas [110]
3 years ago
7

– 18q+10q+16q–3= – 19

Mathematics
2 answers:
sergeinik [125]3 years ago
8 0

Step-by-step explanation:

  • – 18q+10q+16q–3= – 19
  • 8q=-19
  • q=-19/8
  • q=-2.375

dimaraw [331]3 years ago
3 0

Answer:

q = -2

Step-by-step explanation:

-18q + 10q + 16q - 3 = -19

~Combine like terms

8q - 3 = -19

~Add 3 to both sides

8q = -16

~Divide 8 to both sides

q = -2

Best of Luck!

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26.69 square meters

Step-by-step explanation:

A=r(3.14)

A= 8.5(3.14)

A=26.69 meters

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3 years ago
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3 years ago
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4 0
2 years ago
a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was
Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

<em>Let </em>\bar X<em> = sample average time spent studying</em>

The z-score probability distribution for sample mean is given by;

          Z = \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }  ~ N(0,1)

where, \mu = population mean hours spent studying = 25 hours

            \sigma = standard deviation = 15 hours

            n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < \bar X < 30 hours)

    P(29 hours < \bar X < 30 hours) = P(\bar X < 30 hours) - P(\bar X \leq 29 hours)

      

    P(\bar X < 30 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } < \frac{ 30-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z < 2) = 0.97725

    P(\bar X \leq 29 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } \leq \frac{ 29-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z \leq 1.60) = 0.94520

                                                                    

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.</em>

Therefore, P(29 hours < \bar X < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0
3 years ago
A new truck sells for $28,000depreciates (decreases in value) 13% each year. What is the Decay Factor, b?
sleet_krkn [62]
Not sure
te rate is r
for decay, it is
A=P(1-r)^t
r=rate
P=present amount
dunno what b is
but I guess
A=28000(1-0.13)^t
A=28000(0.87)^t
I would guess the decay factor would be 0.87
3 0
3 years ago
Read 2 more answers
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