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djverab [1.8K]
3 years ago
6

A _______ is invested by managers in a diversity of stocks, bonds, and other securities. 

Mathematics
2 answers:
Scorpion4ik [409]3 years ago
8 0
A C. MUTUAL FUND is invested by managers in a diversity of stocks, bonds, and other securities.

A mutual fund is an investment vehicle where a pool of funds are collected from numerous investors and are invested in securities like stocks, bonds, money market instruments, and similar assets. It is operated by money managers, who invest the fund capital on the investors behalf.

A. series EE bond - simply a type of bond
B. preferred stock - company shares with special options
D. promissory note - a note that promises to pay a debt
OLEGan [10]3 years ago
3 0

A mutual fund is invested by managers in a diversity of stocks, bonds, and other securities.


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8 minutes

Step-by-step explanation:

Leo has already run 1 mile, leaving him with 4 more to finish the race. At 2 minutes a mile (a ridiculously impressive pace, considering the world record mile time is just a little under 3:45), it will then take him 4 x 2 = 8 minutes to cross the finish line.

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Find the total surface area of the following prism <br> 5cm <br> 6cm<br> 8cm<br> 4cm
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The total surface area of the given prism is 152 square centimeters.

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2 years ago
A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \
Yanka [14]

Answer:

a) P(A)=0.875

b) \text{P(A or B)}=0.875

c) \text{P((not A)  or B  or (not C))}=0.625

Step-by-step explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

S={HHH,HHT,HTT,HTH,TTT,TTH,THH,THT}

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

P(A)=\frac{n(A)}{n(S)}

P(A)=\frac{7}{8}

P(A)=0.875

b) P(A or B)

Using formula,

\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}

\text{P(A or B)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{\text{n(A and B)}}{n(S)}

\text{P(A or B)}=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}

\text{P(A or B)}=\frac{7}{8}

\text{P(A or B)}=0.875

(c) P((not A)  or B  or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

\text{P((not A)  or B  or (not C))}=\frac{5}{8}

\text{P((not A)  or B  or (not C))}=0.625

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3 years ago
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-95/5=-19

70/-2=-35

-18 \times 10=-180

27 \times -3=-81

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