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lana66690 [7]
3 years ago
13

4.1 is ten times as much as____ 9.2 is ten times as much as___

Mathematics
2 answers:
Setler79 [48]3 years ago
6 0

4.1 is ten times as much as 0.41

9.2 is ten times as much as 0.92

Arada [10]3 years ago
5 0

Answer:

4.1 is 10 x 0.41

9.2 is 10x .92

Can I have brainliest?

Step-by-step explanation:

You might be interested in
George Johnson recently inherited a large sum of money; he wants to use a portion of this money to set up a trust fund for his t
nikdorinn [45]

Answer:

a. B+S = 1

0.06B+0.1S \geq 0.075

B \geq 0.3

Objective function:

R=0.06B+0.1S

b) See Attached picture

30% in bonds and 70% in stocks

Step-by-step explanation:

a.

In order to solve the first part of the problem we need to take into account that the problem wants us to determine the percentage that should be allocated to each of the possible investment alternatives. In that case, the sum of the percentages must be equal to 1 (which means that we will have 100 of the trust fund)

so that gives us our first restriction for the problem.

B+S=1

Next, the problem tells us that the projected returns over the life of the investments are 6% for the bond fund and 10% for the stock fund. It also states that he wants to select a mix that will enable him to obtain a total return of at least 7.5%, so we can take this information and get the second restriction from it:

0.06B+0.1S \geq 0.075

Next, the problem tells us that whatever portion of the inheritance he finally decides to commit to the trust fund, he wants to invest at least 30% of that amount in the bond fund, so that's where the last restriction comes from:

B \geq 0.3

now, the idea is to optimize the investment, this is get the greatest amount of money out of the trust fund, so the objective function is:

R=0.06B+0.1S

which represents the return on the investment.

b)

For part b we can start by graphing each of the restrictions:

B+S = 1

This will be a single line, you can draw the line by setting B=0 first so you get:

0+S=1

S=1

So the first point to plot will be (0,1)

next, we can set s=0 to get:

B+0=1

B=1

so the second point to plot will be (1,0)

so you can plot the two points and connect them with a straight line. This is the green line on the uploaded graph.

Next we can graph the second restriction:

0.06B+0.1S \geq 0.075

we can use the same procedure we used for the previous graph, in this case the points would be:

(0 , 0.75) and (1.25, 0)

and again connect the two points with a straight line. Next we need to decide which region of the graph to shade for which we can pick two arbitrary points on each side of the line, for example we can pick:

(0,0) and (2,2) and see which one makes the inequality true:

for (0,0) we get:

0.06B+0.1S \geq 0.075

0.06(0)+0.1(0) \geq 0.075

0 \geq 0.075

Which is false, therefore we need to shade the other region of the graph:

for (2,2) we get:

0.06B+0.1S \geq 0.075

0.06(2)+0.1(2) \geq 0.075

0.32 \geq 0.075

Which is true, so we shade the region of the graph that contains that point. (see red graph)

now we graph the third restriction:

B \geq 0.3

In order to graph this third restriction we just need to draw a vertical line at B=0.3 and shade everything to the right of that line. (Blue graph)

Now, we can analyze the graph, in this case we need to locate the points where the green line crosses the red and the blue line which gives us the following coordinates:

(0.3, 0.7) and (0.625, 0.375)

these two points can be found by setting the first restriction equal to each of the other two restrictions if you are to do it algebraically. If you are using a graphing device, you can directly read them from the graphs.

So once we got those points, we can see which one gives us the greatest percentage of return.

let's test the first point (0.3, 0.7)

R=0.06B+0.1S

R=0.06(0.3)+0.1(0.7)

R=0.088

so this distribution gives us 8.8% in return, let's test the second point:

(0.625, 0.375)

R=0.06B+0.1S

R=0.06(0.625)+0.1(0.375)

R=0.075

so this distribution gives us 7.5% in return.

In this case the best distribution for us is 30% in bonds and 70% in the stock fund to get a return of 8.8%

6 0
3 years ago
What is the Answer to -14.5 + (-1.24)
Bingel [31]
The answer would be -15.74
4 0
3 years ago
Read 2 more answers
Simplify the ratio 24:81
gizmo_the_mogwai [7]

Answer:

8 to 27

Step-by-step explanation:

3 0
2 years ago
Julian has to read 4 articles for school. He has 8 nights to read them. He decides to read the same number of articles each nigh
daser333 [38]

<u><em>Answer:</em></u>

a. He will have to read half an article per night

b. Fraction of the reading assignment read each night = \frac{0.5}{4}=\frac{1}{8}

<u><em>Explanation:</em></u>

<u>Part a:</u>

<u>We are given that:</u>

number of articles to read = 4 articles

number of nights = 8 nights

To get the number of articles that he should read per night, we will simply divide the the number of articles by the number of nights

<u>Therefore:</u>

number of articles to read per night = \frac{4}{8}=\frac{1}{2} articles per night

<u>Part b:</u>

Now, we know that he will read half an article each night from a total of 4 articles

To get the fraction of the reading assignment read each night, we will divide the number of articles read each night by the total number of assignments

<u>Therefore:</u>

Fraction of the reading assignment read each night = \frac{0.5}{4}=\frac{1}{8}

Hope this helps :)

5 0
3 years ago
There are 6 red marbles 4 blue marbles to yellow marbles. Sabrina picks 2 marbles without putting the first one back what is the
Yanka [14]

Answer:

P = 3/11

Step-by-step explanation:

There are:

6 red marbles

4 blue marbles

2 yellow marbles.

So there are a total of 12 marbles in the bag, such that each one has the same probability of being randomly picked.

We want to find the probability that the first pick is not yellow (so it can be either blue or red), and the second pick is a blue marble.

Now there are two cases.

The first marble is red and the second blue

the first marble is blue and the second blue

We need to find the probability in each case.

The probability that the first marble is red is equal to the quotient between the number of red marbles (6) and the total number of marbles (12)

p = 6/12

Now, for the second draw, we need a blue marble, the probability in this case is the quotient between the number of blue marbles (4) and the total number of marbles in the bag (now there are 11, because one is already taken)

q = 4/11

The joint probability is equal to the product of the individual probabilities, then:

P₁ = p*q = (6/12)*(4/11) = 2/11

And for the other case, we draw two blue marbles:

For the first draw, the probability is:

p = 4/12

And for the second draw, now there are 3 blue marbles and 11 total marbles, then the probability is:

q = 3/11

The joint probability is:

P₂ = (4/12)*(3/11) = 1/11

The total probability is then:

P = P₁ + P₂ = 2/11 + 1/11 = 3/11

4 0
3 years ago
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