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Mashcka [7]
3 years ago
6

Pls help is easyyyyy geometry

Mathematics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer:

A = 30 unit^2

Step-by-step explanation:

To find the area of the triangle

A = 1/2 b*h where b is the length of the base and h is the height

b = 12 and h =5

A = 1/2 (12*5)

A = 1/2 *60

A = 30 unit^2

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research suggests that the probability of a certain fuse malfunctioning increases exponentially as the concentration of an impur
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y=3(257,959)^x[tex]  where x is the impurity &#10;concentration and y, given as a percent, is the probability of the fuse &#10;malfunctioning.\\Then, the probability of the fuse malfunctioning for an impurity concentration of 0.17 is given by [tex]y=3(257,959)^{0.17}=3(8.316941)=24.95

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What is the aswer?

Step-by-step explanation:

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3 years ago
If f(x) = 5x, what is f^-1(x)?<br> F(x) = -5x<br> of(x)-<br> F(x) = 5*<br> O F(x) = 5x
Oksana_A [137]

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3 0
3 years ago
At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more
I am Lyosha [343]

Answer:

The probability is  P( X \le 4 ) = 0.0054

Step-by-step explanation:

From the question we are told that

   The percentage that are on time is  p =  0.76

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Generally the percentage that are not on time is

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The  probability that no more than 4 of them were on time is mathematically represented as

        P( X \le 4 ) =  P(1 ) +  P(2) + P(3) +  P(4)

=>     P( X \le 4 ) =  \left n } \atop {}} \right.C_1 p^{1}  q^{n- 1} +   \left n } \atop {}} \right.C_2p^{2}  q^{n- 2} +  \left n } \atop {}} \right.C_3 p^{3}  q^{n- 3}  +  \left n } \atop {}} \right.C_4 p^{4}  q^{n- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{11- 1} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{11- 2} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{11- 3}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{11- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{10} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{9} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{8}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{7}

= \frac{11! }{ 10! 1!}  (0.76)^{1}  (0.24)^{10} +   \frac{11!}{9! 2!}  (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!}  (0.76)^{3}  (0.24)^{8}  + \frac{11!}{7!4!}  (0.76)^{4}  (0.24)^{7}

P( X \le 4 ) = 0.0054

4 0
3 years ago
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