MATHEMATICS HIGH SCHOOL

Say that the economy is in an expansion, causing wages to increase by six percent. If you previously received a monthly salary of $1,375.00, what will your new monthly salary be? Round all dollar values to the nearest cent.

Answers

Answer 1

Answer:

1457.50 dollars or 1458 dollars rounded

Answer 2

Answer:

Answer:

1457.50

Step-by-step explanation

Took the tester

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Does the equation x^2+y=1 define y as a function of x

Solve for n
3n-2=2n
N=

Answers

Answer:

Hey there!

3n-2=2n

n-2=0

n=2

Let me know if this helps :)

3n - 2 = 2n

n =?

3n - 2n = 2

n = 2 ✔️

An airline flies between 6 airports. How many different flight paths can it offer, if it lands three times and never stays in the same place twice?

Answers

Answer:

the airline can offer 20 different flight paths under the given conditions.

Step-by-step explanation:

If the airline has 6 airports and the plane lands three times without staying in the same place twice, this is essentially a permutation problem. You want to find the number of ways to arrange 3 distinct airports out of 6.

This can be calculated using the formula for permutations of "n" items taken "r" at a time:

nPr = n! / (n - r)!

Where "n" is the total number of items (airports in this case) and "r" is the number of items to be arranged (3 landings in this case), and "!" denotes factorial.

So, in your case, the calculation would be:

6P3 = 6! / (6 - 3)!

= (6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)

= 120 / 6

= 20

So, the airline can offer 20 different flight paths under the given conditions.

Sam's hourly wage increased from $10 an hour to $20 an hour. What is thepercent increase of her hourly wage? (Round to the nearest whole number)
A) 150%
B) 200%
C) 100%
D) 50%

Answers

I think it’s 100%......

Which of the following numbers would you multiply -3/4 by to equal 1?
@Sbcardinals

Answers

Just multiply -3/4 by -4/3 :)

You have to multiply -3/4 by a negative number to make it positive and the 3/4 cancels out into 1 when multiplied by 4/3. Hope this helped!

Multiply 2 1/2 x -4 3/5
Plz help

Answers

its -11.5 (or 1/2, you can write it like that too) if i'm not mistaken

Answer:

Step-by-step explanation:

2 1/2* -4 3/5=

5/2*-23/5

-23/2 or -11 1/2

Among the 10 most popular sports, men include competition-type sports - pool and billiards, basketball, and softball - whereas women include aerobics, running, hiking, and calisthenics. However, the top recreational activity for men was still the relaxing sport of fishing, with 41% of those surveyed indicating that they had fished during the year. Suppose 180 randomly selected men are asked whether they had fished in the past year. Suppose 180 randomly selected men are asked whether they has fished in the past year.a. What is the probability that fewer than 50 had fished?
b. What is the probability that between 50 and 75 (inclusive) had fished?
c. If the 180 men selected for the interview were selected by the marketing department of a sporting-goods company based on information obtained from their mailing lists, what would you conclude about the reliability of their survey results?

Answers

Answer:

(a) The probability that fewer than 50 had fished is 0.0002.

(b) The probability that between 50 and 75 (inclusive) had fished is 0.6026.

Step-by-step explanation:

Let X = number of men who had fished during the year.

The probability of the random variable X is, p = 0.41.

A random sample of n = 180 men are selected.

The random variable X follows a Binomial distribution with parameters n = 180 and p = 0.41.

A Normal approximation to Binomial is applied when the following conditions are met,

np ≥ 10
n(1 - p) ≥ 10

Check:

$np=180*0.41=73.8>10\nn(1-p)=180* (1-0.41)=63.72>10$

Thus, the distribution of x can be approximated by a Normal distribution with:

Mean = $np=180*0.41=73.8$

Standard deviation = $√(np(1-p))=√(180*0.41*(1-0.41))=6.599$

(a)

Compute the probability that fewer than 50 had fished as follows:

$P(X<50)=P((X-\mu)/(\sigma)>(50-73.8)/(6.599))\n=P(Z<-3.61)\n=1-P(Z<3.61)\n=1-0.9998\n=0.0002$

Thus, the probability that fewer than 50 had fished is 0.0002.

(b)

Compute the probability that between 50 and 75 (inclusive) had fished as follows:

$P(50\leq X\leq 75)=P(49.5<X<75.5)\n=P((49.5-73.8)/(6.599)<(X-\mu)/(\sigma)<(75.5-73.8)/(6.599))\n=P(-3.68<Z<0.26)\n=P(Z<0.26)-P(Z<-3.68)\n=0.6026-0\n=0.6026$

Thus, the probability that between 50 and 75 (inclusive) had fished is 0.6026.

(c)

If the sample of 810 men are selected from mailing list then it is highly probable that the sample is not a representative of the true population, i.e. sports men.

Because the people interested in sports are less likely to be interested in fishing.

Thus, it could be concluded that the survey results are not reliable.

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