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kap26 [50]
2 years ago
7

Help!!! evaluate peicewise functions

Mathematics
1 answer:
aksik [14]2 years ago
4 0

Answer:

9

Step-by-step explanation:

You choose the middle one because 3<(4)<10 makes sense where as the other ones don't when you plug in 4. Then you solve the equation which is x^2-7 and then you get 16-7 which is 9.

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What is the answer to -6&lt;21
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Step-by-step explanation:

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How many different values can an algebraic expression like 5+(x+2)
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3 years ago
A multiple-choice standard test contains total of 25 questions, each with four answers. Assume that a student just guesses on ea
weqwewe [10]

Answer:

a)  8*88*10⁻⁶ ( 0.00088 %)

b) 0.2137 (21.37%)

Step-by-step explanation:

if the test contains 25 questions and each questions is independent of the others, then the random variable X= answer "x" questions correctly , has a binomial probability distribution. Then

P(X=x)= n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of questions= 25

p= probability of getting a question right = 1/4

then

a) P(x=n) = p^n = (1/4)²⁵ = 8*88*10⁻⁶ ( 0.00088 %)

b) P(x<5)= F(5)

where F(x) is the cumulative binomial probability distribution- Then from tables

P(x<5)= F(5)= 0.2137 (21.37%)

7 0
3 years ago
1.Arsenic-74 is used to locate brain tumors. It has a half-life of 17.5 days. 90 mg wereused in a procedure. Write an equation t
Sophie [7]

1)\text{   }N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}

2) 5.625 mg will be left

Explanation:

1) Half-life = 17.5 days

initial amount of Arsenic-74 = 90 mg

To get the equation, we will use the equation of half-life:

\begin{gathered} N_t\text{ = N}_0(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}} \\ where\text{ N}_t\text{ = amount remaining} \\ N_0\text{ = initial amount} \\ t_{\frac{1}{2}\text{ }}\text{ = half-life} \end{gathered}N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}

2) we need to find the remaining amount of Arsenic-74 after 70 days

t = 70

\begin{gathered} N_t=\text{ 90\lparen}\frac{1}{2})^{\frac{70}{17.5}} \\ N_t\text{ = 5.625 mg} \end{gathered}

So after 70 days, 5.625 mg will be left

4 0
1 year ago
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