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3 years ago

Help me please I need help

Mathematics

1 answer:

balandron [24]3 years ago

8 0

Answer:

I have solved the question in the picture in two different ways : simplification and factorisation I hope you find it useful bye:)

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I. Direction: Write the following statement as equations. Write your answers

Bingel [31]

Answer:

1. 4x - 29=11

2. (3x - 4) (2) = 46

3. 3/5x + 8 = 50

4. The difference of 51 and the sum of x and 5.

5. The product of 3 and y minus eight.

Step-by-step explanation:

4 0

3 years ago

I need help on these please

sleet_krkn [62]

Answer:

its the third answer, fifth, and first. Im not sure tho.

Step-by-step explanation:

if it isnt right im so sorry T^T

5 0

3 years ago

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}$

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

$P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483$

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

5 0

3 years ago

Four more than 3 times a number t written asfour

Orlov [11]

Four more than 3 times a number t

3t+4

If you are telling me to write the variable t as 4- then:

3(4)+4

12+4

16

Hope this helps!
~cupcake

5 0

3 years ago

Read 2 more answers

What can you say about the sum of consecutive odd numbers starting with 1? that is, evaluate 1, 1c3, 1c3c5, 1c3c5c7, and so on,

Mice21 [21]

Let us add consecutive odd numbers and try to find any relationship.

1. 1

2. 1+3 = 4 ( square of 2 i.e $2^{2}$ )

3. 1+3+5 = 9 ( $3^{2}$ )

4. 1+3+5+7 = 16 ( $4^{2}$ )

5. 1+3+5+7+9 = 25 ( $5^{2}$ )

6. 1+3+5+7+9+11 = 36 ( $6^{2}$ )

7. 1+3+5+7+9+11+13 = 49 ( $7^{2}$ )

If we notice, the sum of the consecutive odd integers in each case is equal to the square of the place where it lies. For example, the sum of numbers in seventh place is equal to $7^{2}$ . The sum of the numbers in the fifth line is equal to $5^{2}$ .

4 0

3 years ago