If two numbers on a fraction are negative, does that make the fraction positive

41% of the students at an art college want to be a graphic designer about what fraction of the students want to be a graphic des

IgorC [24]

I think the answer is 41 over 100

4 0

4 years ago

WILL GIVE BRAINLIEST. please help

kondor19780726 [428]

Answer:

x = 11.4 m

y = 21 m

man's shadow = 9.6 m

Step-by-step explanation:

measure of third angle must be 50 degrees (180 - (40 + 90))

you can take tan50° = 15/y and 'y' will equal approx. 21

to find 'x' you can take tan21° = (21-x) /25 and 21 - x = 25 · tan21°

21 - x = 9.6

-x = -11.4

x = 11.4

man's shadow is the difference of 21 and 11.4, which is 9.6

4 0

3 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

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

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

3 years ago

A hiker sets out to climb a mountain with a 4,000 ft vertical ascent as shown in the diagram below. Ifthe angle of incline, a =

Ne4ueva [31]

hi fxgxgicgicigci log oh

6 0

3 years ago

Neptune is an average distance of 4.5 × 10^9 km from the Sun. Estimate the length of the Neptunian year using the fact that the

zlopas [31]

Answer:

164.32 earth year

Step-by-step explanation:

distance of Neptune, Rn = 4.5 x 10^9 km

distance of earth, Re = 1.5 x 10^8 km

time period of earth, Te = 1 year

let the time period of Neptune is Tn.

According to the Kepler's third law

T² ∝ R³

$\left ( \frac{T_{n}}{T_{e}} \right )^{2}=\left ( \frac{R_{n}}{R_{e}} \right )^{3}$

$\left ( \frac{T_{n}}{1} \right )^{2}=\left ( \frac{4.5\times10^{9}}}{1.5\times10^{8}}} \right )^{3}$

Tn = 164.32 earth years

Thus, the neptune year is equal to 164.32 earth year.

Step-by-step explanation:

3 0

3 years ago