Answer: $7.6502
Step-by-step explanation:
Grape cost $1.19 per pound. 3.16 pounds are bought. The total amount paid for grape will be:
= $1.19 × 3.16
= $3.7604
Peaches cost $1.29 per pound. 1.35 pounds are bought. The total amount paid for peaches will be:
= $1.29 × 1.35
= $1.7415
Peers cost $0.99 per pound. 2.17 pounds are bought. The total amount paid for pears will be:
= $0.99 × 2.17
= $2.1483
Total bill paid will be:
= $3.7604 + $1.7415 + $2.1483
= $7.6502
Since, it is given that the pentagon ABCDE is congruent to pentagon FGHIJ.
By being congruent, the corresponding angles and sides are equal.
Therefore,
,
,
,
,
.
By using 
So, 
Therefore, the measure of angle 1 is 100 degrees.
So, Option B is the correct answer.
Is it 79 sets of ten or 79 tenths
Answer: The expected value of this game is 2/3
Step-by-step explanation:
Give that
If it's black, you lose a point. If it's red, you gain a point.
And then you can stop at any time. But you should never stop when you are losing because that can guarantee 0 by drawing all the cards.
Assuming you should stop after three cards when you are +2.
The only question is whether to draw if you are +1 on the first draw.
If you draw red first, You have 1/3 chance of drawing red again and this will give you +2 points
1/3 chance of drawing two blacks and earn zero point, chance of drawing black-red and earn +1. This gives +1, so it doesn't matter whether you draw or not.
From the beginning, If you draw red (probability 1/2 you end +1. If you draw black and then draw two reds (probability 1/6 you end +1) Otherwise you break even with probability 1/3. Overall, the value is 2/3
Let
. Then differentiating, we get

We approximate
at
with the tangent line,

The
-intercept for this approximation will be our next approximation for the root,

Repeat this process. Approximate
at
.

Then

Once more. Approximate
at
.

Then

Compare this to the actual root of
, which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.