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ivanzaharov [21]
3 years ago
13

I need help setting it up and solving it but mostly setting it all up please and thank you ?!!!i need help asap!!!!!

Mathematics
1 answer:
professor190 [17]3 years ago
8 0
This is the work that I did. Hope it helps

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This straightedge and compass constructions shows a quadrilateral ABCD. is ABCD a rhombus
dezoksy [38]

Answer:

Is this topic symmetry?

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Incorrect

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Help! how do i answer this
navik [9.2K]
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7 0
3 years ago
On March 17th, Pinny Sporting Goods accepted a credit card purchase for $900. The credit card company charges a 2% service fee.
Vera_Pavlovna [14]

Answer:

A : debit to Service Charge Expense of $18.

Step-by-step explanation:

First, one should calculate the value, in dollars of the credit card service fee

Fee = 2% x $900 = $18.

Since this fee is classified as "money coming out" of the account, the entry to a record of this transaction should include a debit to Service Charge Expense of $18.

The correct answer is alternative A.

*The whole purchase amount ($900) should be credited and the service expense should be debited, unlike alternative C, which merges both entries into one.

6 0
3 years ago
A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student
Gwar [14]

Answer:

B) 0.283

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

25% of the students drive themselves to school.

This means that p = 0.25

Class of 18 students

This means that n = 18

What would be the probability that at least 6 students drive themselves to school?

This is

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006

P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034

P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096

P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17

P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213

P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282

Closest option is B, just a small rounding difference.

4 0
3 years ago
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