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3 years ago

1 Suppose that $67,000 is invested at 55% interest, compounded quarterly.

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

5 0

There is missing data here. We have the rate and principal amount but no value given in terms of years.

The formula needed is A = p(1 + (r/n))^(nt).

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Annette [7]

6E1 divided by 2E-1 equals 300

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3 years ago

Slope Intercept form of a Line.

valkas [14]

Answer:

Step-by-step explanation:

Slope: -3

y-intercept: -2 (y-intercept is where a line cross the y-axis, vertical line)

Pencil on the y-intercept, three down and one to the right, then trace the line from -2 on the y-axis.

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2 years ago

A bicycle wheel has a diameter of 63centermeter.the wheel.during a journey the wheel makes a 510revolation turn.how many meter d

olga55 [171]

63 × 510 = 32130

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3 years ago

Write answer in slope intercept form (3,-8); slope= -3

Slav-nsk [51]

Answer:

$y=-3x-8$

Step-by-step explanation:

The slope intercept form is $y=mx+b$ .

The " $mx$ " is the slope; which in this case is -3. So $-3x$ .

The " $b$ " is the y-intercept, which is this case is the -8 because it is the 2nd number in the coordinate point.

This is why the answer is $y=-3x-8$ .

4 0

3 years ago

Read 2 more answers

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago