All categories

2 years ago

I have no clue lolI accidently put supplementary so ignore that.

Mathematics

1 answer:

Anika [276]2 years ago

7 0

Answer:

5. Vertical

6. 133°

Step-by-step explanation:

These are vertical angles; Vertical angles are congruent. Therefore, x = 133°.

You might be interested in

Round to the nearest whole.<br> 782.36

ratelena [41]

The answer would be 782 because you would round down.

6 0

2 years ago

Read 2 more answers

A metal manufacturing company reduces the length of a 12.5-inch cylindrical rod, after which the rod measures 7.8 inches. How ma

gizmo_the_mogwai [7]

Answer:

4.7 inches

Step-by-step explanation:

The original is 12.5, and it was shortened to 7.8. So take the original, subtract the result, and you will get the answer.

12.5-7.8=4.7

7 0

2 years ago

Read 2 more answers

Find the value of a in this isosceles trapezoid. D. 11 A. 7 B. 15 C. 4

UNO [17]

Answer:

Step-by-step explanation:

Because 15-4=11

and both angles are equal

4 0

2 years ago

A particular brand of dishwasher soap is sold in three sizes: 25 oz, 40 oz, and 60 oz. Twenty percent of all purchasers select a

ss7ja [257]

Answer:

Please find solution in the picture attached

Step-by-step explanation:

4 0

2 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

2 years ago

I have no clue lolI accidently put supplementary so ignore that.​

I have no clue lolI accidently put supplementary so ignore that.