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ehidna [41]
3 years ago
11

Courtney and Malik are buying a rug to fit in a

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
4 0
Where’s the rest of the problem
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What is the inequality shown?
ale4655 [162]

Answer:

1 is inequality

replace the value were _2.

8 0
3 years ago
Factor completely 2x^2 − 2x − 40. 2(x − 5)(x 4) (2x − 10)(x 4) (x − 5)(2x 8) 2(x − 4)(x 5)
Zina [86]
2x^2 - 2x - 40 = 2x^2 -10x + 8x - 40 = 2x(x - 5) + 8(x - 5)

= (x-5)(2x+8) = 2(x-5)(x+4)
3 0
3 years ago
143=7+8(3-7x) solve and justify
Llana [10]
Justify means plug answer in and verify

so distribute 8(3-7x)=24-56x

143=7+24-56x
143=31-56x
minus 31 both sides
112=-56x
divide both sides by -56
-2=x

justify

plug it back
143=7+8(3-7x)
143=7+8(3-7(-2))
143=7+8(3-(-14))
143=7+8(3+14)
143=7+8(17)
143=7+136
143=143

true

x=-2

7 0
3 years ago
School lunch costs $1.25 per day. If there are 21 school days
prohojiy [21]

Answer:

$26.25

Step-by-step explanation:

1.25 per day

21 days

1.25 × 21 = 26.25

3 0
2 years ago
Read 2 more answers
A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), <em>p</em> = 0.52

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        =P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

=P(X

Thus, the probability that the flight has empty seats is 0.4625.

4 0
3 years ago
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