Old
9.99×55
=549.45
New
10.68×55
=587.4
((10.68÷9.99)−1)×100
=6.9%
Answer: After 18.05 minutes, the temperature of steel becomes 100 degrees.
Step-by-step explanation:
Since we have given that
Initial temperature = 2500
At t = 0,
we get that

After 2 minutes, the temperature of the steel is 1500 degrees.
so, it becomes,

So, We need to find the number of minutes when the temperature of steel would be 100 degrees.
So, it becomes,

Hence, after 18.05 minutes, the temperature of steel becomes 100 degrees.
This is a two step problem. First, you need to find the difference (or increase) in the rent using subtraction.
616-560 = $56 increase
Now we need to figure what percent 56 is to the original amount (560)
We voice it like this:
56 is to 560 as x is to 100.
The equation looks like this:
56/560 - x/100
Cross multiply to cancel out the denominators.
56 x 100 = 560 x
5600 = 560 x
x = 10 percent.
Check it!
What is 10 percent of 560?
560 x .1 = $56
Was that your increase? ... yes.
Answer: 125.80 ft
Step-by-step explanation:
Asuming the described situation is as shown in the figure below, we need to find the distance
between the kite and Jacks house, but first we need to find the
,
and then
.
How?
We will use trigonometry, especifically the trigonometric functions sine and cosine:
For
:
(1)
Where
is the opposite side to the angle and
the hypotenuse.
Isolating
:
(2)
For
:
(3)
Where
is the adjacent side to the angle.
Isolating
:
(4)
Finding
:
(5)

(6)
Now that we have found these values, we have to work with a bigger triangle, where the hypotenuse is the distance between the kite and Jack's house
and the sides are the values calculated in (4) and (6).
So, in this case we will use the <u>Pithagorean theorem</u>:
(7)
Isolating
and writing with the known values:
(8)
(9)
This is the distance between the kite and the house
Answer:
26.11% of the test scores during the past year exceeded 83.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8. This means that
.
Approximately what percentatge of the test scores during the past year exceeded 83?
This is 1 subtracted by the pvalue of Z when
. So:



has a pvalue of 0.7389.
This means that 1-0.7389 = 0.2611 = 26.11% of the test scores during the past year exceeded 83.