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alisha [4.7K]
3 years ago
11

What does w=11 plus negative 7

Mathematics
2 answers:
nata0808 [166]3 years ago
8 0

Answer:

w = 4

Step-by-step explanation:

w = 11 + (-7)

w = 11 - 7

w = 4

kow [346]3 years ago
5 0

Answer:

w=4

Step-by-step explanation:

adding a negative is like subtracting so w=11-7

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if x and y are two overlapping subsets of a universal set U, write the relation between n(U),n(X U Y), and n(xūy).​
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Answer:

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n(U) = n(X U Y) + n(xūy).

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Esther pays $355 per month for 5 years for a car. She made a down payment of $2,500. If the loan costs 7.1% per year compounded
oksano4ka [1.4K]
The cash price of the car includes the amount of the loan plus the amount of the down payment
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Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
Pv the amount of the loan ?
PMT payment per month 355
R interest rate 0.071
K compounded monthly 12
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×5))÷(0.071÷12))
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5 0
3 years ago
A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi
SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

= \frac{20!}{(20-5)! 5!)}

= \frac{20*19*18*17*16}{5*4*3*2*1}

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

= \frac{20}{320} = \frac{1}{19}

P(A5) = ⁸C₅ / ²⁰C₅

= \frac{56}{15540} = \frac{7}{1938}

P(A4) = ⁷C₄ / ²⁰C₄

= \frac{35}{4845} = \frac{7}{969}

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0
3 years ago
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