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mel-nik [20]
3 years ago
14

Fred Salmon purchased six $1,000 bonds at 92. The bonds pay 6.5%, what is the cost of the bonds

Mathematics
2 answers:
soldi70 [24.7K]3 years ago
4 0
The cost of the bond
6×1,000×0.92=5,520
IgorLugansk [536]3 years ago
4 0

Answer:

$5,520.00

Step-by-step explanation:

Fred Salmon purchased six $1,000 bonds at 92. The bonds pay 6.5%, what is the cost of the bonds

Lets look at the question

What was the cost of the bonds?

He bought 6 $1,000 bonds, so 6 times 1,000 = 6,000. Nat the age of 92. Which means 6,000 x 0.92 to get $5,520.00

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If x+y=16 and x-y=-2, what is the value of xy?
Fantom [35]

Answer:

Y=7 and

X=9

Step-by-step explanation:

X+Y=16

X=2+Y

2+Y+Y=16

2+2Y=16

2Y=16-2

2Y=14

2Y/2=14/2

Y=7

X =2+Y

So X=9

8 0
11 months ago
If a rope 36 feet long is cut into two pieces in such a way that one piece is twice as long as the other piece, how long must th
SVEN [57.7K]
24 feet long :)
All I did was I divided the 36 into 3 equal parts of 12, then doubled the 12 to make that longer piece 24 feet. This 24 foot piece would be twice as long as the other leftover 12 foot piece since 12x2=24
4 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.9degreesf and a standard deviation of 0.67
trasher [3.6K]

Answer:

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = 0.67

n = number of samples = 103

From the information given, the population standard deviation is unknown hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 103 - 1 = 102

Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.01/2 = 0.005

the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995

Looking at the t distribution table,

z = 2.6249

Margin of error = 2.6249 × 0.67/√103

= 0.173

Confidence interval = 98.6 ± 0.173

This suggests that the mean body temperature could very possibly be

98.6degrees°F.

3 0
3 years ago
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6 0
3 years ago
Read 2 more answers
Somebody help me with this plsss it for geometry
ICE Princess25 [194]
Use an online math calculator for more accurate answers just plug in the variables
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2 years ago
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