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Novosadov [1.4K]
3 years ago
6

What is 11.5+y in word phase

Mathematics
1 answer:
Ksivusya [100]3 years ago
6 0
Eleven and five tenths plus another number
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Trees planted by a landscaping firm have a 90% one-year survival rate, If they plant 16 trees in a park, what is the following p
irina1246 [14]

Answer:

Pr(all will survive in one year) = 0.35 to 2 decimal places

Pr(at least 8 will survive) = 0.39 to 2 decimal places.

Step-by-step explanation:

Probability of a tree surviving one year is 0.9. While not surviving probability is 1 - 0.9 = 0.1

This is obtained from the 90℅ surviving rate for one year planted by the landscaping firm.

Probability that if 10 trees are planted, all will survive for one year is:

0.9x0.9x0.9x0.9x0.9x0.9x0.9x0.9x0.9x0.9 = 0.9^10 = 0.349

Probability that at least 8 will survive in one year can be expressed as

8 survived x 2 not survived +(or) 9 survived x 1 not survived +(or) 10 survived

= [(0.9^8)x(0.1^2) + (0.9^9)x(0.1^1) + (0.9^10)] = 0.391

Step-by-step explanation:

6 0
3 years ago
Write down inequalities,that are satisfied by these sets of integers between -10 and 10
Anastaziya [24]

Answer:

Below

Step-by-step explanation:

Notice that x is between 10 and -10 but takes only the values that are integers.

The inequalities:

● we can write an inequality that includes all these values.

● -10 《 x 《 10

This is a possible inequality

Multiply both sides by 2 and you will get a new one:

● -20 《 2x 《 20

You can multiply it by any number to generate a new inequality.

Or you can add or substract any number.

8 0
3 years ago
Describe el perimetro
mrs_skeptik [129]
<span>Tomás compró 350 metros de malla galvanizada para vallar el perímetro del lote</span>
3 0
3 years ago
Traverion wants to find the answer to the question “How much money does the average college professor make?” He surveys 50 profe
Grace [21]

Answer: The results do not represent the population because the sample does not represent professors from different areas.


Step-by-step explanation:

Since Traverion wants to find the answer to the question “How much money does the average college professor make?”

He only surveys 50 professors from the university in his hometown.

Although he should survey professors from different areas but he only perform his survey only a specific university in his hometown.

Therefore, the survey is biased as the sample size is too small.

Hence, The results do not represent the population because the sample does not represent professors from different areas.

4 0
3 years ago
Read 2 more answers
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
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