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2 years ago

Choose the solution to this inequality. 72≥b+95

Mathematics

1 answer:

Travka [436]2 years ago

7 0

Answer:

False

72 is not greater then or equal to 95 on its own, let alone when another quantity is added to 95

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Jerry has a credit card debt of $15,600 that he would like to reduce by applying $8500 of his inheritance money to the balance

aev [14]

Answer:

Jerry would still be in debt by 7100

Step-by-step explanation:

Just subtract 8500 from 15600

6 0

2 years ago

Use the Parent function and Function 2 to choose the best statement comparing the graphs to each other.

kogti [31]

Answer:

the answer is A&d

Step-by-step explanation:

I took the quiz and got 100

3 0

3 years ago

Round 5.9496 to the nearest thousandth

Evgesh-ka [11]

Answer:

5.95

Step-by-step explanation:

that is because 96 is rounded too 100

so its 5.9500 = 5.95

4 0

3 years ago

Read 2 more answers

Can someone help me?<br> (if you can show your work that would be nice).

lubasha [3.4K]

Answer:

1. x1=15.75.

2. 24/48=0.5

3. 24/25=0.96

Step-by-step explanation:

1. 15.75/18=0.875

2. you multiply 48 times 0.5 to find 24.

3. multiply 0.96 with 25 to get 24.

3 0

3 years ago

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago