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4 years ago

Larry received the bank statement below.

Mathematics

2 answers:

SIZIF [17.4K]4 years ago

6 0

Check 318 was not cashed or deposited is the answer.

rewona [7]4 years ago

6 0

Answer:

Option D. check 318 was not cashed or deposited.

Step-by-step explanation:

First we will check the statement transactions

Total with-drawls done = 58.29 + 75.40 + 121.57 + 33.52 = $288.78

Total deposits = 250 + 250 = $500

Opening balance was $750 so as per our calculations closing balance should be 750 + 500 - 288.78 = $961.22 which matches with bank's balance of the statement.

But Larry's checkbook has a balance of $922.63 which different from the bank's balance.

This clearly states that check 318 was not cashed or deposited which is making a difference.

Option D is the answer.

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Points A and B have coordinates A(-4, 2) and B(3,-6). Find the coordinates of point P, the weighted average of points A and B, i

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Answer:

C) P(1, -26/7)

Step-by-step explanation:

You want the weighted average of A(-4, 2) and B(3,-6) using weights 2 and 5, respectively.

<h3>Weighted average</h3>

Each of the values is multiplied by the corresponding weight, and the sum of those products is divided by the total of the weights to obtain the weighted average.

P = (2A +5B)/(2+5)

P = (2(-4, 2) +5(3, -6))/7 = (-8 +15, 4 -30)/7 = (7, -26)/7

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The coordinates of P are (1, -26/7).

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2 years ago

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4 years ago

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Step-by-step explanation:

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3 years ago

PLS PLS HELP!!! I NEED THIS DONE PLS!!!

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The domain and the range are 0 ≤ ∅ ≤ 2π and -1 ≤ cos(∅) ≤ 1, respectively.

<h3>The domain and the range</h3>

The domain is the set of input values i.e. the ∅ values.

On the table, we can see that the ∅ value is from 0 to 2π.

This means that the domain is 0 ≤ ∅ ≤ 2π

On the other hand, the range is the set of output values i.e. the cos(∅) values.

On the table, we can see that the cos(∅) value is from -1 to 1.

This means that the range is -1 ≤ cos(∅) ≤ 1

Hence, the domain and the range are 0 ≤ ∅ ≤ 2π and -1 ≤ cos(∅) ≤ 1, respectively.

<h3>The points on a graph</h3>

See attachment 1

<h3>The graph of the function</h3>

See attachment 2

Read more about domain and range at:

brainly.com/question/2264373

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2 years ago

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 120 customer orders to f

Aliun [14]

Answer:

a. P(X = 0) = 0.02586

b. $\mathbf{P(X \leq 2 ) =0.2879}$

c. $\mathbf{P(X \leq 5 ) =0.8387}$

Step-by-step explanation:

From the given information:

a. If the manufacturer stocks 120 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X = 0)=(^{120}_{0}) (0.03)^0 (1-0.03)^{n-0}$

$P(X = 0)=\dfrac{120!}{0!(120-0)!} (0.03)^0 (1-0.03)^{n-0}$

P(X = 0) = 1 × 1 ( 0.97)¹²⁰ ⁻ ⁰

P(X = 0) = 0.02586

b. ) If the manufacturer stocks 122 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 2 ) = [ P(X=0) + P(X =1) + P(X = 2) ]$

$P(X \leq 2 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2}]$ $P(X \leq 2 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120}]$

$P(X \leq 2 ) = [(1 \times 1 \times 0.02433 )+(122 \times (0.03) \times 0.025083)+(7381 \times 9 \times 10^{-4} \times 0.02586)]$

$\mathbf{P(X \leq 2 ) =0.2879}$

(c) If the manufacturer stocks 125 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 5 ) = [ P(X=0) + P(X =1) + P(X = 2) +P(X = 3)+P(X = 4)+ P(X = 5) ]$

$P(X \leq 5 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2} + (^{122}_{3})(0.03)^3 (0.97)^{122-3} + (^{122}_{4})(0.03)^4 (0.97)^{122-4}+ (^{122}_{5})(0.03)^5 (0.97)^{122-5}]$ $P(X \leq 5 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120} + \dfrac{122!}{3!(122-3)! }*(0.03)^3(0.97)^{122-3)}+ \dfrac{122!}{4!(122-4)! }*(0.03)^4(0.97)^{122-4)} +\dfrac{122!}{5!(122-5)! } *(0.03)^5(0.97)^{122-5)}]$

$\mathbf{P(X \leq 5 ) =0.8387}$

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3 years ago