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3 years ago

Show work pls i need help

Mathematics

1 answer:

Anestetic [448]3 years ago

6 0

This is the answers:
20: 6.9/7
21: 6.36/6
22: 6.24/6
23: 9.1/9
24: 4.6/5
25: 5.8/6
26: 9.3/9
27: 4.12/5

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Simplify <br> 1+sinx/cosx+cosx/1+sinx

Dmitry [639]

$\frac{1 + sin(x)}{cos(x)} + \frac{cos(x)}{1 + sin(x)}$
$\frac{[1 + sin(x)][1 +sin(x)]}{cos(x)[1 + sin(x)]} + \frac{cos(x)[cos(x)]}{cos(x)[1 + sin(x)]}$
$\frac{1 + 2sin(x) + sin^{2}(x)}{cos(x)[1 + sin(x)]} + \frac{cos^{2}(x)}{cos(x)[1 + sin(x)]}$
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6 0

3 years ago

A photograph 8 cm by 11 cm will be framed. The combined area of the frame and photograph will be 180 cm^2. Algebraically determi

SCORPION-xisa [38]

Answer:

12 cm x 15 cm

Step-by-step explanation:

if the picture frame thickness is x,

then the frame dimension would be (8+2x) cm by (11+2x) cm

area is (8+2x)(11+2x)=180

solve for x, x= 2, -23/2

x is thickness, so should be >0, so x=2

frame dimension: 12 x 16

8 0

2 years ago

3x to the 2 power -17+2x+5-x

rewona [7]

Answer: $=3x^{2^{x-12}}$

Step-by-step explanation:

$3x^{2^{-17+2x+5-x}}$

Simplify:

$=3x^{2^{x-12}}$

5 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

Find the area of the shaded region. Round to the nearest hundreth where necessary.

Jet001 [13]

Answer:

192.42 m²

Step-by-step explanation:

big circle

10.5 ² x pi =346.3605

inner circle

7 ² x pi = 153.9380

346.3605 - 153.9380 = 192.4225 (192.42)

3 0

3 years ago