Find a tenth of three quarters of $800

The adult hummingbird weights 0.056 ounces. This is three times lighter than the worlds largest butterfly. How much does the but

Likurg_2 [28]

The butterfly weighs .168, as far as the scientific notation, it's 1.68x10^2

8 0

3 years ago

3x+2y-7x+33y^6<br> x=7<br> y=18

andrew-mc [135]

Answer:

math is was invented just so school is torture

Step-by-step explanation:

5 0

3 years ago

What would be the resulting equation in the graph of the parent function y=x^2 was shifted 2 units to the right and 3 units up?

Musya8 [376]

Answer:

(X-2)^2+3

Step-by-step explanation:

For rightward movement subtract from x (leftward movement -add), then add to the overall equation to shift it upwards (subtract for downwards)

5 0

3 years ago

The amount of time it takes to recover physiologically from a certain kind of sudden noise is found to be normally distributed w

Taya2010 [7]

Answer:

a) 2.5%

b) 97.5%

c) 16%

d) 84%

e) 50%

f) 50%

g) 84%

h) 16%

i) 97.5%

j) 2.5%

Step-by-step explanation:

The empirical rule, or the rule of 50%-34%-14%, states that:

In a normally distributed stat with mean $\mu$ and standard deviation $\sigma$ ...

a) 2.5% of the scores are going to be above $\mu + 2\sigma$

b) 13.5% of the scores are going to be above $\mu + \sigma$ and below $\mu + 2\sigma$ .

c) 34% of the scores are going to be above $\mu$ and below $\mu + \sigma$

d) 34% of the scores are going to be above $\mu - \sigma$ and below $\mu$

e) 13.5% of the scores are going to be above $\mu - 2\sigma$ and below $\mu - \sigma$

f) 2.5% of the scores are going to be below $\mu - 2\sigma$

In this problem

We have that $\mu = 80s$ and $\sigmma = 10s$

So:

(a) above 100, (b) below 100

$100 = \mu + 2\sigma = 80 + 2*10$

So 2.5% of the scores are going to be above 100, and the other 97.5% is going to be below 100

c) above 90, (d) below 90

$90 = \mu + \sigma = 80 + 10$

So 13.5% of the scores are going to be above 90 and below 100, and 2.5% of the scores are going to be above 100. So 13.5% + 2.5% = 16% of the scores are going to be above 90 and the other 84% is going to be below 90

(e) above 80, (f) below 80

80 is the mean, so approximately 50% percent of the scores are going to be above 80 and 50% are going to be below 70%.

(g) above 70, (h) below 70

$70 = \mu - \sigma = 80 - 10$

34% of the scores are going to be above 70 and below 80, and other 50% percent of the scores are going to be above 80. So in all, 84% of the scores are going to be above 70. The other 16% of the scores are going to be below 70.

(i) above 60, and (j) below 60

$60= \mu - 2\sigma = 80 - 2*10$