of 183 + 198 is less than 300. Is

Jamal's

answer reasonable? Why or

why not

Both addends are

des to 200.

Answer:

Answer:

Jamal's answer isn't reasonable because the sum of 183 and 198 is 381, which is way more than 300 and nowhere less than 300.

Step-by-step explanation:

Jamal makes an assertion that the sum of 183 and 198 is less than 300.

We are to check if Jamal's answer is reasonable or not.

183 + 198 = 381 > 300

The sum of the two numbers, 381, is evidently not less than 300, hence, Jamal's answer isn't reasonable because it is downright wrong.

Hope this Helps!!!

There are 5 slices of pepperoni pizza, 1 slice of sausage pizzá, and 3 slices of cheese pizza left at the pizza party. Without looking, Amy took a slice of pizza, ate it, and then took another slice. What is the probability of Amy eating two slices of cheese pizza?

A study is being conducted to compare the average training time for two groups of airport security personnel: those who work for the federal government and those employed by private security companies. From a random sample of 12 government-employed security personnel, average training time was 72 hours, with a sample standard deviation of 8 hours. In a random sample of 16 privately employed security personnel, training time was 65.4 hours, with a sample standard deviation of 12.3 hours. Assume that training time for each group is normally distributed. Use the following notations:μ1: The mean training time for the population of airport security personnelemployed by the federal government.μ2: The mean training time for the population of airport security personnelemployed by private security companies.The goal of the statistical analysis is to determine whether the sample data support the hypothesis that average training time for government-employed security personnel is higher than those employed by private security companies.1. What is the null hypothesis H0?Select one:a. μ1- μ2 <= 0b. μ1- μ2 < 0c. μ1- μ2 =/ 0d. μ1- μ2 > 02. What is the alternative hypothesis Ha?Select one:a. μ1- μ2 > 0b. μ1- μ2 <= 0c. μ1- μ2 = 0d. μ1- μ2 >= 0

1011For questions 6-8, determine the number of solutions.Type your answers as one, none, or many (no capital letters)64(x + 5) - 2 = 18 + 4x

What's the inverse operation of division? A. Multiplication B. All mathematical operations C. Subtraction D. Division has no inverse operation.

A child flies a kite at a height of 150 ft, the wind carrying the kite horizontally away from the child at a rate of 78 ft/sec. How fast must the child let out the string when the kite is 390 ft away from the child?

A study is being conducted to compare the average training time for two groups of airport security personnel: those who work for the federal government and those employed by private security companies. From a random sample of 12 government-employed security personnel, average training time was 72 hours, with a sample standard deviation of 8 hours. In a random sample of 16 privately employed security personnel, training time was 65.4 hours, with a sample standard deviation of 12.3 hours. Assume that training time for each group is normally distributed. Use the following notations:μ1: The mean training time for the population of airport security personnelemployed by the federal government.μ2: The mean training time for the population of airport security personnelemployed by private security companies.The goal of the statistical analysis is to determine whether the sample data support the hypothesis that average training time for government-employed security personnel is higher than those employed by private security companies.1. What is the null hypothesis H0?Select one:a. μ1- μ2 <= 0b. μ1- μ2 < 0c. μ1- μ2 =/ 0d. μ1- μ2 > 02. What is the alternative hypothesis Ha?Select one:a. μ1- μ2 > 0b. μ1- μ2 <= 0c. μ1- μ2 = 0d. μ1- μ2 >= 0

1011For questions 6-8, determine the number of solutions.Type your answers as one, none, or many (no capital letters)64(x + 5) - 2 = 18 + 4x

What's the inverse operation of division? A. Multiplication B. All mathematical operations C. Subtraction D. Division has no inverse operation.

A child flies a kite at a height of 150 ft, the wind carrying the kite horizontally away from the child at a rate of 78 ft/sec. How fast must the child let out the string when the kite is 390 ft away from the child?

Answer:

Step-by-step explanation:

The original price of the wind chimes at the home improvement store is $w.

A customer signed up for a free membership card and received a 5% discount off the price. The value of the discount is

5/100 × w = 0.05w

The discounted price would be

w - 0.05w = 0.95w

Sales tax of 6% was applied after the discount. The amount of sales tax applied would be

6/100 × 0.95w = 0.057w

The algebraic expression to represent the final price of the wind chime is

0.95w + 0.057w

= 1.007w

The answer is 8 ihinghkki

**Answer:**

P(B)=0.30

**Step-by-step explanation:**

Out of 1000 Voters, 30% favor Jones.

Event S=Favors Jones on First Trial

Event B=S occurs on Second Trial

P(S)=0.30

P(S')=1-0.30=0.70

Event B could occur in two ways

- The first two trials are a success
- The first trial is a failure and the second trial is a success.

Therefore,

P(B)=P(SS)+P(S'S)

=(0.3X0.3)+(0.7X0.3)

=0.09+0.21

=0.3

Therefore, the probability of event B(that event S occurs on the second trial), **P(B)=0.30.**

**Answer: irdk the awnser but F on that tumor**

**Answer:**

y-8=1(x-4)

**Step-by-step explanation:**

I used the equations y-y/x-x and y-__y__=__m__(x-__x__)

y=-8=1(x-4) hope that helped!

**Answer:**

The number of times organism B's population is larger than organism A's population after 8 days is 32 times

**Step-by-step explanation:**

The population of organism A doubles every day, geometrically as follows

a, a·r, a·r²

Where;

r = 2

The population after 5 days, is therefore;

Pₐ₅ = = 32·a

The virus cuts the population in half for three days as follows;

The first of ta·2⁵ he three days = 32/2 = 16·a

The second of the three days = 16/2 = 8·a

After the third day, Pₐ = 8/2 = 8·a

The population growth of organism B is the same as the initial growth of organism A, therefore, the population, P₈ of organism B after 8 days is given as follows;

P₈ = a·2⁸ = 256·a

Therefore, the number of times organism B's population is larger than organism A's population after 8 days is P₈/Pₐ = 256·a/8·a = 32 times

Which gives, the number of times organism B's population is larger than organism A's population after 8 days is 32 times.

Organism A's population at the end of 5 days is 2^5. After 5 days, a virus cuts it in half for 3 days. Organism B's population at the end of 8 days is 2^8. To find the difference, subtract organism A's population from organism B's population.

Organism A's population doubles every day for 5 days, so the population at the end of 5 days is 25. After 5 days, a virus cuts the population in half for 3 days, so we need to find (25) * (2-1)3. Using the rule of exponents, we can rewrite this expression as (25+(-1*3)), which simplifies to 2-4.

Organism B's population grows at the same rate but is not infected with the virus. After 8 days, the population is 28.

To find out how much larger organism B's population is than organism A's population, we need to subtract the population of organism A from organism B. So, 28 - 2-4 is the answer.

#SPJ3