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3 years ago

1) $400 interest is earned on a principal of $2,000 at a simple interest rate of 5%

Mathematics

1 answer:

Kitty [74]3 years ago

3 0

Answer:

4 years

Step-by-step explanation:

I = $ 400

R = 5%

P =$ 2000

I = Prt

400 = 2000 * $\frac{5}{100}$ * t

400 = 20 * 5 * t

400 = 100 t

$\frac{400}{100}=t$

t = 4 years

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2 years ago

The height of a triangle is 8 cm more than the base. If the area is 24 cm2, find the height and base of the triangle.

Elden [556K]

Answer:

height = 12 cm

base length = 4 cm

Step-by-step explanation:

area of a triangle

base length × height / 2

x = height

y = base length

x = y + 8

24 = y × (y + 8) / 2

48 = y × (y + 8) = y² + 8y

squared equation

y² + 8y - 48 = 0

solution

y = (-b ± sqrt(b² - 4ac))/(2a)

a = 1

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c = -48

y = (-8 ± sqrt(64 - 4×-48))/2 = (-8 ± sqrt(64 + 192))/2 =

= (-8 ± sqrt(256))/2 = (-8 ± 16)/2 = -4 ± 8

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but a negative base length did not make any sense, so only y = 4 remains.

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3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

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

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

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3 years ago