Total calories = 56
Amount of calories in plum = 32
Remaining Calories = 56 - 32 = 24
Amount of calories with one strawberry = 4
No. of strawberries he ate = 24/4 = 6 strawberries
I hope it is helpful:D
Answer:
D
Step-by-step explanation:
We are given that:
And we want to find the value of tan(2<em>x</em>).
Note that since <em>x</em> is between π/2 and π, it is in QII.
In QII, cosine and tangent are negative and only sine is positive.
We can rewrite our expression as:
Using double angle identities:
Since cosine relates the ratio of the adjacent side to the hypotenuse and we are given that cos(<em>x</em>) = -1/3, this means that our adjacent side is one and our hypotenuse is three (we can ignore the negative). Using this information, find the opposite side:
So, our adjacent side is 1, our opposite side is 2√2, and our hypotenuse is 3.
From the above information, substitute in appropriate values. And since <em>x</em> is in QII, cosine and tangent will be negative while sine will be positive. Hence:
<h2>
</h2>
Simplify:
Evaluate:
The final answer is positive, so we can eliminate A and B.
We can simplify D to:
So, our answer is D.
Amount after 15 years = 6700( 1 + 0.046/2)^(2*15) = 13,253.90
9514 1404 393
Answer:
3) y = -1
5) x = -14
Step-by-step explanation:
The first step is to recognize that the equation describes a vertical line in problem 3 and a horizontal line in problem 5. The perpendicular to a vertical line is a horizontal line, and vice versa.
__
3. To make the desired horizontal line go through the point (-8, -1) the y-value of the line must match that of the point:
y = -1
__
5. To make the desired vertical line go through the point (-14, 81), the x-value of the line must match that of the point:
x = -14
Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
