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3 years ago

Please help!!!!!!!!!!!

Mathematics

1 answer:

Effectus [21]3 years ago

4 0

I feel like it's everything except for choices b and c. I hope this helps, but I would read them again.

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54 thousands 6 hundreds 12 tens – 15 thousands 8 hundreds 26 tens = ________(1 Point)

Afina-wow [57]

Answer:

38,660

Step-by-step explanation:

54 thousands 6 hundreds 12 tens – 15 thousands 8 hundreds 26 tens

54 thousands = 54000

6 hundreds = 600

12 ten = 12*10 = 120

54 thousands 6 hundreds 12 tens = 54000+600+120 = 54720

15 thousands = 15000

8 hundreds = 800

26 tens = 26*10 = 260

15 thousands 8 hundreds 26 tens = 15000+800+260 = 16060

Thus,

54 thousands 6 hundreds 12 tens – 15 thousands 8 hundreds 26 tens

= 54720- 16060 = 38,660 (Answer)

3 0

3 years ago

Please Help!!

Vlad [161]

Answer:

1. 1.28

2. 55

3. +

Step-by-step explanation:

1. 0.043 x 10 {3}

= 0.043 x (10 x 3)

= 0.043 x 30

= 1.28

2. 5 + (9) (6) - 4

= 5 + 9 x 6 - 4

= 5 + 54 - 4

= 59 - 4

= 55

3. Which of the these operations should be completed last when solvingan equation?

a. x

b. +

c. ()

d.÷

b. +

6 0

3 years ago

Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of 14 flight

Bezzdna [24]

Answer:

a. p=1.000

b. p=0.2924

c. p=0.7358

d. No

Step-by-step explanation:

a. This problem satisfies all the criteria for a binomial experiment expressed as:

$P(X=x){n\choose x}p^x(1-p)^{n-x}$

-Given that p=0.85, n=14, the probability that exactly all 14 were on time is calculated as:

$P(X\geq 1)=1-P(X=0)\\\\=1-{12\choose 0}0.85^0(1-0.85)^{12}\\\\=1-1.297\times 10^{-10}\\\\=1.0000$

Hence, the probability that all 12 flights are on time is 1.0000

b. Given that n=12, and p=0.85

-The probability that exactly 10 flights are on time is calculated as;

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\={12\choose 10}0.85^{10}(1-0.85)^{2}\\\\=0.2924$

Hence, the probability that exactly 10 flights are on time is 0.2924

c. Given that n=12, and p=0.85

-The probability that more of 10 or more flights are on time:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 10)=P(X=10)+P(X=11)+P(X=12)\\\\={12\choose 10}0.85^{10}(0.15)^{2}+{12\choose 11}0.85^{11}(0.15)^{1}+{12\choose 12}0.85^{12}(0.15)^{0}\\\\=0.2924+0.3012+0.1422\\\\=0.7358$

Hence, the probability of 10+ flights being on time is 0.7358

d. We first find the mean of the distribution:

$\mu=E(X)=0.85\times 14\\\\=11.9$

#We then find the probability of 11+=0.3012+0.1422=0.4434

-We compare the expectation to the probability of 11+ flights being on time.

No. Since the probability $P(X\geq 10)$ =0.4434 < that the expectation, 11.9, it is not unusual for 11+ flights to be on time.

*I have used a sample size of n=12 since there are two separate n values:

5 0

3 years ago

Previous 20 Item 20 The width of a rectangular room is 60% of its length. The length of the room is 10 feet. What is the area of

zimovet [89]

Answer: A=60 square feet

Explanation: w=10*0.6=6 ft.

A=lw

A=10*6=60 square feet.

4 0

3 years ago

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Westkost [7]

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