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3 years ago

Betty has a checking account and a non-interest bearing savings account. The function C(x), shown below, represents her checking

Mathematics

2 answers:

Tems11 [23]3 years ago

6 0

Answer:

T(x) or total account balance = C(x) + S(x)

Step-by-step explanation:

where = C(x) = -2401 - 140 + 1,400

and

S(x) = 4503 + 100

This implies that:

C(x) = 3,661

S(x) = 4,603

Therefore, T(x) 3,661 + 4,603 = 8,264

T(x) is the total account balance or the addition of the balances of checking and saving accounts.

C(x) is the checking account balance.

S(x) is the saving account balance.

Mamont248 [21]3 years ago

6 0

Answer

(On attachment)

Step-by-step explanation:

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The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that $\mu = 74, \sigma = 15$

Question a:

Sample of 36 means that $n = 36, s = \frac{15}{\sqrt{36}} = 2.5$

This probability is 1 subtracted by the pvalue of Z when X = 78. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{78 - 74}{2.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that $n = 150, s = \frac{15}{\sqrt{150}} = 1.2247$

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

$Z = \frac{X - \mu}{s}$

$Z = \frac{77 - 74}{1.2274}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 71

$Z = \frac{X - \mu}{s}$

$Z = \frac{71 - 74}{1.2274}$

$Z = -2.45$

$Z = -2.45$ has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that $n = 219, s = \frac{15}{\sqrt{219}} = 1.0136$

This probability is the pvalue of Z when X = 74.2. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{74.2 - 74}{1.0136}$

$Z = 0.2$

$Z = 0.2$ has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0

3 years ago

1 Point

Gre4nikov [31]

Answer:

C. The length is 6 times the width.

Step-by-step explanation:

Hope it helps you in your learning process.

5 0

3 years ago

A 9,000-lb load is suspended from the roof in a shopping mall with a 16-ft-long solid aluminum rod. The modulus of elasticity of

Westkost [7]

Answer:

Step-by-step explanation:

Given:

elongation, x = 0.50 in

Force, f = 9000 lb

Young modulus, E = 10,000,000 psi

Maximum Stress, Sm = 30000 psi

Length, L = 16 ft

Converting ft to in,

12 in = 1 ft

=16 × 12 = 192 in

Young modulus, E = stress/strain

Stress = force/area, A

Strain = elongation, x/Length, L

E = f × L/A × E

1 × 10^7 = stress/(0.5/16)

= 26041.7 psi

Minimum stress = 26041.7 psi

Maximum stress = 30,000 psi

Stress = force/area

Area = 9000/26041.7

= 0.3456 in^2

Stress = force/area

Area = 9000/30000

= 0.3 in^2

Using minimum area of 0.3 in^2,

A = (pi/4)(d^2)

0.3 in^2 = (pi/4)(d^2)

d = 0.618 inches

diameter, d = 0.618 inches

7 0

3 years ago

Expression of "three fifths of a number b subtracted from thrice the same number" into mathematical symbol

liubo4ka [24]

Answer:

3b - $\frac{3}{5}$ b

Step-by-step explanation:

Let the number be "b"

So the expression would be:

3b - $\frac{3}{5}$ b

6 0

2 years ago

What would the total area be??

timurjin [86]

Answer:

1,560,000 mm²

Step-by-step explanation:

A= $\frac{(1600*600)}{2}$ This would make shape A's area 480,000.

B= $\frac{(600*600)}{2}$ This would make shape B's area 180,000.

C= $\frac{(1000*600)}{2}$ This would make shape C's area 300,000.

D= 1000×600 This wouls make D's area 600,000.

Now you add up all of the areas;

480,000+180,000+300,000+600,000= 1,560,000 mm²

6 0

3 years ago

Read 2 more answers