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Leokris [45]
3 years ago
9

You have invested $5000 in stocks and bonds. The interest you earned on the stocks was 7%. The interest you earned on the bonds

was 14%. At the end of the year, you earned a total of $500 from your investments. How much money did you invest in the stocks and how much money did you invest in the bonds?
Mathematics
1 answer:
Sphinxa [80]3 years ago
8 0
Money invested S + B = $5000 - - - - (a)
Money earned can be represented in S & B as follows:
0.07*S + 0.14*B =$500 - - - - (b)

Equations a & b are two equations with two unknowns.

From equation a:
S = 5000 - B
Replace S in equation b
0.07*(5000 - B) + 0.14*B =$500
0.07B = 500 - 350 = $150
B = 150/0.07 = $2142.86
S = 5000 - 2142.86 = $2857.14

Good luck
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What is the equation of a parabola if the vertex is (1, 4) and the directrix is located at y = 7?
Oksana_A [137]

According to the vertex and the directrix of the given parabola, the equation is:

y = \frac{3}{4}(x - 1)^2 + 4

<h3>What is the equation of a parabola given it’s vertex?</h3>

The equation of a quadratic function, of vertex (h,k), is given by:

y = a(x - h)^2 + k

In which a is the leading coefficient.

The directrix is at y = k + 4a.

In this problem, the vertex is (1,4), hence:

h = 1, k = 4

The directrix is at y = 7, hence:

4 + 4a = 7

a = \frac{3}{4}

Hence, the equation is:

y = a(x - h)^2 + k

y = \frac{3}{4}(x - 1)^2 + 4

More can be learned about the equation of a parabola at brainly.com/question/26144898

4 0
2 years ago
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l
Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal(\mu=4.5,\sigma^{2} =0.70^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 4.5 minutes

           \sigma = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X \leq 4.50 min)

    P(X < 5.30 min) = P( \frac{X-\mu}{\sigma} < \frac{5.30-4.5}{0.7} ) = P(Z < 1.14) = 0.8729

    P(X \leq 4.50 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.5-4.5}{0.7} ) = P(Z \leq 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

    P(X > 5.30 min) = P( \frac{X-\mu}{\sigma} > \frac{5.30-4.5}{0.7} ) = P(Z > 1.14) = 1 - P(Z \leq 1.14)

                                                              = 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 5.30 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 5.30 min) = P( \frac{X-\mu}{\sigma} \leq \frac{5.30-4.5}{0.7} ) = P(Z \leq 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 4.00 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 4.00 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.0-4.5}{0.7} ) = P(Z \leq -0.71) = 1 - P(Z < 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

            P(X > x) = 0.05            {where x is the required time}

            P( \frac{X-\mu}{\sigma} > \frac{x-4.5}{0.7} ) = 0.05

            P(Z > \frac{x-4.5}{0.7} ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

                      \frac{x-4.5}{0.7}=1.645

                      {x-4.5}{}=1.645 \times 0.7

                       x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0
3 years ago
Please help with this due asap
yuradex [85]
A.
6/9 + 6/9 = 12/9
Hope this helps!!!
7 0
2 years ago
Read 2 more answers
Find the sum of 16+22+28+...+112
Elodia [21]
Im not 100 percent positive that this is correct but I think the answer is 
184
6 0
3 years ago
Can someone help me with this ONE question? Write an equation.. and solve for X
matrenka [14]

Answer:

Equation: 720=(8x-1)+110+(5x+36)+116+(7x+19)+(6x-2)

x=17

Step-by-step explanation:

A 6 sided polygon is 720° degrees

so

720=(8x-1)+110+(5x+36)+116+(7x+19)+(6x-2)

Simplify

720=(8x+5x+7x+6x)+(−1+110+36+116+19+−2)

Combine like terms:

720=26x+278

flip equation:

26x+278=720

subtract 278 from both sides

26x=442

divide by 26 both sides:

x=17

Hope this helps.

7 0
2 years ago
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