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3 years ago

Together, Sala and Terry saved $10.00. Sala. saved D dollars. Write an expression for the amount Terry saved.

Mathematics

2 answers:

givi [52]3 years ago

6 0

Answer:

T = 10 - S

Step-by-step explanation:

The only useful information is that they saved $10 together. We are just going to add the money Sala and Terry saved to equal 10.

S + T = 10

They want an expression for how much Terry saved so we can solve for T.

T = 10 - S

Best of Luck!

vodka [1.7K]3 years ago

3 0

Let Terry Saved be T dollars .

The Expression will be

$\\ \rm\longmapsto Sala+Terry=\$10.00$

$\\ \rm\longmapsto D+T=\$10.00$

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Answer: 72 feet

Step-by-step explanation: 1 yard is equal to 3 feet, so if there are 24 yards, we can multiply 24 by 3. This can be shown using the following formula: y · 3 = f (where y = number of yards and f = feet.) So:

24 · 3 = 72

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3 years ago

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A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

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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 = the length of the calls, in minutes.

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 = 0.3729.

(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 = 0.1271

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 = 0.1109.

(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 = 0.745.

(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.

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3 years ago

Stan has made a $125.30 monthly deposit into an account that pays 1.5% interest, compounded monthly, for 35 years. he would now

Andrew [12]

The annuity of the monthly deposit into an account that pays 1.5% interest, compounded monthly, for 35 years is $333.71

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An annuity is a series of payments made at equal period of time.

future value = annuity x [(1 + i)ⁿ - 1] / i

annuity = $125.30

i = 1.5% / 12 = 0.00125

n = 35 years x 12 months = 420

future value = $125.30 x [(1 + 0.00125)⁴²⁰ - 1] / 0.00125

future value = $69,156.049 ≈ $69,156.05

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i = 1.5% / 12 = 0.00125

n = 20 years x 12 months = 240

present value = $69,156.05

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annuity = $86.45 / 0.25904

= $333.71

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2 years ago

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Answer:

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Step-by-step explanation:

Since the rocket is launched from the ground, So = 0 and S(t) = 0

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Where g acceleration due to gravity = -4.9m/s^2. and

initial velocity = 39.2 m/a

0 = -4.9t2 + 39.2t

4.9t = 39.2

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Substitute t in the model equation

S(t) = -49(8^2) + 3.92(8) + So

Let S(t) =0

0 = - 313.6 + 31.36 + So

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The equation that can be used to model the height of the rocket after t seconds will be:

S(t) = -4.9t^2 + Vot + 282.24

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3 years ago

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A package of 24 pencils costs 2.88$. at this rate how much would 30 pencils cost?

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3 years ago

Together, Sala and Terry saved $10.00. Sala. saved D dollars. Write an expression for the amount Terry saved.​

Together, Sala and Terry saved $10.00. Sala. saved D dollars. Write an expression for the amount Terry saved.