Combine these radicals. -3sqrt(of81)+sqrt(of16)

Answer:

The mass of the water removed is 66.67 kg

Step-by-step explanation:

Given data

The solids blended with sugar make up 10%.

Final product achieved = 15% product.

Final product contains = 15% sugar solids.

The initial product weighs = 200 kg

Solution

Here, we find the needed sugar amount as follows:

The amount of sugar required = 200 ×

Thus, the needed sugar = 20 kg

We recognize that total solids = product solids + sugar solids...............1

Also, the initial product equals the final product.

So,

0.20 × 200kg= 0.30 × total mass of the final product

The mass of the final product = 133.33 kg

However,

the final product contains 15% product solids and 15% sugar solids, which means that the amount of product solids equals the sugar solids.

Thus, the mass of water removed = 200 kg - 133.33 kg.

The mass of the water removed is 66.67 kg.

Step-by-step explanation:

∑⁴ₙ₌₁ -144 (½)ⁿ⁻¹

This represents a finite geometric series where n equals 4, a₁ is -144, and r is ½.

S = a₁ (1 − rⁿ) / (1 − r)

S = -144 (1 − (½)⁴) / (1 − ½)

S = -270

If you wish to calculate the infinite sum (n = ∞):

S = a₁ / (1 − r)

S = -144 / (1 − ½)

S = -288

∠ROT=160°

∠SOT=100°

Now we calculate ∠SOR as follows: ∠SOR = ∠ROT - ∠SOT

∠SOR = 160° - 100°

∠SOR = 60°

It is stated that the angles ROQ, QOS, and POT all have the same measure.

Thus, ∠SOQ + ∠QOR = 60°

Since ∠SOQ equals ∠QOR, we can express this as:

2∠SOQ = 60°

From which we find ∠SOQ = 60° ÷ 2

∠SOQ = 30°

Also, ∠POT = ∠SOQ = ∠ROQ = 30°

In the absence of a specific question posed, below are the potential inquiries along with their respective answers:

P(fewer than 4 tosses)
= P(one toss) + P(two tosses) + P(three tosses)
= (3/4) + (3/4)(1/4) + (3/4)(1/4)^2
= 0.984375


Expected value
= 1 / p
= 1 / (3/4)
= 4 / 3

Variance
= (1 - p) / p^2
= (1 - (3/4)) / (3/4)^2
= (1/4) / (9/16)
= 4 / 9

Standard deviation
= sqrt(Variance)
= sqrt(4 / 9)
= 2 / 3