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Tresset [83]
2 years ago
5

With this diagram, what could be the values of c and d?

Mathematics
1 answer:
Sedaia [141]2 years ago
7 0

Answer:

idk maybe

Step-by-step explanation:

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Solve the inequality: -6n - 12 >= 3(-n + 12)
Verizon [17]

Answer:

-16 > n or n < -16

Step-by-step explanation:

First, we can write the inequality and distribute.

-6n-12 > 3(-n+12)

-6n-12 > -3n+36

Get rid of a variable first (it makes solving a lot easier). We can add 6n to both sides.

-6n-12 > -3n+36

<u>+6n          +6n</u>

-12 > 3n+36

Now, we can subtract 36 on both sides.

-12 > 3n+36

<u>-36         -36</u>

-48 > 3n

Divide by 3 on both sides.

-48 > 3n

<u>/3       /3</u>

-16 > n or (if it's easier) n < -16

Hope this helps!! Have a wonderful day C:

7 0
2 years ago
How long can late payments stay on your credit report?
ratelena [41]
The exception is if you are 60 days<span> late often which will certainly hurt your scores. Otherwise, one late payment should not cause long term damage. </span>90 days<span>late: This record will damage your credit scores significantly for up to </span>seven years<span>.</span>
5 0
2 years ago
Introduction to complex numbers 5VV
sesenic [268]

Step-by-step explanation:

sqrt(-4) = sqrt(4×-1) = sqrt(4)×sqrt(-1) = 2i

6 0
1 year 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
Will give brainliest, Use properties of exponents to write an equivalent expression.
Sav [38]

Answer:

see attached

Step-by-step explanation:

4 0
2 years ago
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