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iVinArrow [24]
2 years ago
15

Emily spent $55 from her savings account on a new dress explain how to describe the change in emilys savings balance in two diff

erent ways
Mathematics
1 answer:
mote1985 [20]2 years ago
8 0
You can either express the decrease that happened in the account savings by integers. To do this you simply subtract 55$ from the initial amount in the account. This is expressed as:
initial amount - 55$

or, you can express the decrease in the form of percentage by calculating how much is 55$ from the total amount. This is expressed as:
(5/initial amount) x 100
You might be interested in
The information below is the number of daily emergency service calls made by the volunteer ambulance service of Walterboro, Sout
mafiozo [28]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: The number of emergency service calls made by the volunteer ambulance service of Walterboro, South Carolina for 50 days.

a)

In the first column, you have the possible number of emergency calls made, in the second column, you have the observed absolute frequency (fi) for each value of the variable.

To establish the probability distribution you have to calculate the relative frequency (hi) for each value of X.

The formula for the relative frequency is

hi= fi*n where i= 0, 1, 2, 3, 4

X₁= 0

f₁= 8

h₁= 8/50= 0.16

X₂= 1

f₂= 10

h₂= 0.20

X₃= 2

f₃= 22

h₃= 22/50= 0.44

X₄= 3

f₄= 9

h₄= 9/50= 0.18

X₅= 4

f₅= 1

h₅= 1/50= 0.02

(See attachment)

b)

The variable of interest is a cuantitative discrete variable, so this is an example of a driscrete distribution.

Discrete variables are numerical and take certain numbers within its range of definition, contrary to the continuous variables that can take any value within the range of definition of the variable.

Another example of a discrete variable is "the money", If the variable is "I have at most 10 dollars in my wallet", the range of definition goes from 0 dollars to 10 dollars.

c) To calculate the mean when the data is arranged in a frequency table you have to use the following formula:

X[bar]= ∑Xihi= (0*0.16)+(1*0.2)+(2*0.44)+(3*0.18)+(4*0.02)= 1.7

This means that they expect an average of 1.7 calls per day.

d) The formula for the standard deviation is:

S=\sqrt{\frac{1}{n-1}[sumX^2fi-\frac{(sumXfi)^2}{n} ] }

∑X²fi= (0²*8)+(1²*10)+(2²*22)+(3²*9)+(4²*1)=  195

∑Xfi= (0*8)+(1*10)+(2*22)+(3*9)+(4*1)= 85

S=\sqrt{\frac{1}{49}[195-\frac{(85)^2}{50} ] }= 1.015

I hope this helps!

3 0
2 years ago
A bicycle manufacturer has a stock of 50 frames 70 wheels, and 120 reflectors. if each bicycle manufactured requires 4 reflector
Rainbow [258]
30 I think but this is probably a trick question so I'm just gonna assume I'm wrong. 

5 0
2 years ago
Victoria builds a dollhouse in the shape of a right prism, as show below
Snezhnost [94]
Your answer is correct
7 0
2 years ago
Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b
horrorfan [7]

Answer:

a. X\sim N(\mu = 6.1, \sigma = 1.9) b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is z_{1} = (6.4-6.1)/1.9 = 0.1579 and the z-score related to 7 kg is z_{2} = (7-6.1)/1.9 = 0.4737, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0
2 years ago
Use forward and backward difference approximations of O(h) and a centered difference approximation of O(h2) to estimate the firs
strojnjashka [21]

Answer:

Answer has explained below.

Step-by-step explanation:

Consider the function is:

F(x) = 25x3 – 6x2 +7x -88

Differentiate with respect to x, we get

F’(x) = 25. 3x2 – 6.2x + 7

       = 75x2 – 12x +7

At x = 2, we have

F (2) = 25(2)3 – 6(2)2 + 7(2)-88

        =102

And f’(2) = 75(2)2 – 12 (2) +7

               =283

Now, calculate forward divided difference as:

xi + 1 = xi + h

        =2 + 0.25

        =2.25

F (xi + 1) = f (2.25) = 25 (2.25)3 – 6(2.25)2 +7(2.25) -88

                            =182.21

f’(2) = f(2.25) – f(2) / 0.25 = 182.21 – 102 / 0.25

                                             = 320.84

Єt = 283 – 320.8 / 283 = -13.36%

Now calculate backward divided difference:

Xi-1 = xi –h = 2 – 0.25 = 1.75

F(xi-1)= f(1.8) = 25 . (1.8)3 -6 (1.8)2 + 7 (1.8) – 88

                       = 50.96

F’(2) = f(2) – f(1.8) / 0.25 = 102 – 50.96 / 0.25 = 204.16

Єt = 283 – 204.16 / 283 = 27.86%

Finally, centered divided difference is obtain by inserting f(xi+1) and f (xi-1):

F’(2) = f(2.25) – f(1.8) /2 x 0.25 = 320.84 -50.96 / 0.5 = 539.68

Єt = 283 – 539.68 / 283 = -90.7%

6 0
2 years ago
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