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Vesnalui [34]
3 years ago
6

What is a 2 digit number between 20 and 99

Mathematics
2 answers:
Mazyrski [523]3 years ago
5 0

Any number between 20 and 99

PilotLPTM [1.2K]3 years ago
4 0
Any number from 21-98 is between 20 and 99.
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What the nearest hundred thousand for 767074
Law Incorporation [45]
7 is  digit for  hundred thousands and 6  is for ten thousands so the answer is 800,000
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After 3.5 hours pasha had traveled to 17 miles if she travels at a constant speed how far will she have traveled after four hour
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Since we want to know how far she goes every hour to multiply that by 4 to get how far she goes in 4 hours, we do 17/3.5 to get per hour miles = around 4.86. Multiply that by 4 to get around 19.43 miles
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What is the answer? 3(x-5)+2(x-1)
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5 0
3 years ago
Divide £72cm in the ratio 2:3:1 working out please
ale4655 [162]
2:3:1
2+3+1 = 7

£72 / 7 = £10.28

Answer:
£10.28 * 2 = £20.56
£10.28 * 3 = £30.84
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Please make this answer the brainiest!
4 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

   The  sample size is n =  11

   

Generally the percentage that are not on time is

     q =  1- p

     q =  1-  0.76

     q = 0.24

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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