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Sav [38]
2 years ago
10

Richard’s checking account balance was $57.34 at the beginning of the week. During the week, he recorded the transactions below.

Mathematics
1 answer:
sergij07 [2.7K]2 years ago
5 0

Richard’s account balance at the end of the week is $ 132.45

<h3>What is Account Balance ?</h3>

The amount of money in a person's bank account is called the account balance of that person

It is given that

Account balance of Richard’s at the beginning of the week = $57.34

Deposits in the week  : $ 163.75

Expenses in the week : Groceries + Credit card bill + Gas

Total Expenses in the week = 25.37 + 50 + 13.27

                                               = $ 88.64

The Richard’s account balance at the end of the week = Account Balance +Total Deposits - Total Withdrawals

= 57.34 + 163.75 - 88.64

Therefore , Richard’s account balance at the end of the week is $ 132.45.

To know more about Account Balance

brainly.com/question/23271078

#SPJ1

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Answer:

43°

Step-by-step explanation:

(x + 53)° + (x + 74)° + 73° = 180°

(x + 53 + x + 74)° = 180° - 73°

(2x + 127)° = 107°

2x + 127 = 107

2x = 107 - 127

2x = - 20

x = - 20/2

x = - 10

m\angle A = (x +53)\degree \\\\m\angle A = (-10 +53)\degree \\\\m\angle A = 43\degree \\\\

4 0
3 years ago
A coin contains 9 grams of nickel and 16 grams of copper, for a total weight of 25 grams. What percentage of the metal in the co
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X for coin Y for copper
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3 years ago
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Look at photo for question. Do number 3 A-D?​
Digiron [165]

Answer:

A: 6, 12, 18, 24, 30

10, 20, 30, 40

B: 30

C: 5/6= 25/30 and 7/10= 21/30

D: 25/30>21/30 so 5/6>7/10

8 0
3 years ago
Find the value of 15.0 N in pounds. Use the conversions 1slug=14.59kg and 1ft=0.3048m .
Sliva [168]

The value of 15.0 N in pounds = 3.375 lb

<h3><em>Further explanation </em></h3>

The derivative magnitude is a quantity derived from the principal amount

7 main quantities have been determined based on international standards, namely:

  • 1. Length, meters (m)
  • 2. Time, second (s)
  • 3. Mass, kilograms (kg)
  • 4. Temperature, kelvin (K)
  • 5. Light intensity, candela (cd)
  • 6. Electric current, ampere (A)
  • 7. Amount of substance, mol (m)

Newton (N) is one unit of force (F) which is a quantity derived from mass (m), length (l) and time (s)

Newton can be expressed in terms of Kg m / s²

Formula used:

F = m. a

N = kg. m / s²

In the British system there are units of slug, feet, and pounds so that the unit of force becomes:

F = m. a

Lb = slug. ft / s²

The relationship between the two-unit systems above is:

1 kg = 0.0685 slug

1 slug = 14.59 kg

1 newton = 0.225 lb

1 lb = 4,448 newtons

1 ft = 0.3048 m

So if we change 15 N to lb by using slug and ft

15 N = 15 * (1 / 14.59) slug * (1 / 0.3048) ft / s²

15 N = 3,375 slug ft / s ²

15 N = 3,375 lb

or we convert directly:

15 N = 15 * 0.225 lb

15 N = 3,375 lb

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Keywords: slug, Newton, feet, kg, pound

6 0
3 years ago
The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experi- mental conditio
zzz [600]

Answer:

see explaination

Step-by-step explanation:

Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

Mean X-bar = 31.23

SD = 0.689

a)

Null Hypothesis : Xbar = mu

Alternate Hypothesis : Xbar > mu

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 30 ) /(0.689/sqrt(6))

= 4.372

P-value = ~0

Since P-vale < 0.01 , we will reject null hypothesis.

The data suggest that true average stopping ditance exceeds the maximum.

b)

i) SD = 0.65

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

P-value = 0.3859 Answer

ii) SD = 0.65

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

= -2.9

P-value = 0.0037 Answer

c)

i) SD = 0.8

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

= -2.357

P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

d)

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

=> n = 43 Answer

ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

=> sqrt(n) = (0.65*(-1.28)) / (31.23 - 31)

=> n = 13 Answer

4 0
3 years ago
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