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Lelechka [254]
2 years ago
9

7 - r/8 - 3d Variables r=32 and d= 2

Mathematics
1 answer:
satela [25.4K]2 years ago
6 0

The required value if r = 32 and d = 2 is -3

Given the expression

7-\frac{r}{8} - 3d

Substituting the given variables

r = 32

d = 2

The expression becomes

= 7-\frac{32}{8} - 3(2)\\= 7 - 4 - 6\\= 3 - 6\\= -3

Hence the required value is -3

Learn more here: brainly.com/question/17499155

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30

Step-by-step explanation:

2700/90 = 30

27/9 = 3

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2 years ago
Brainliest for right answer!!!
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Let y=x+6 be equation 1...
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\frac{20}{29}Answer:

sin α = \frac{20}{29}

Step-by-step explanation:

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2 years ago
Show by substitution whether the number r is a solution of the corresponding quadratic equation.
Stels [109]

Answer:

It's choice B.

Step-by-step explanation:

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= 121 - 88 - 33

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3 0
2 years ago
businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message
KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: \lambda=\frac{62.7}{24}= 2.6125.

The probability of a random variable can be computed using the formula:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

             =1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

6 0
2 years ago
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