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3 years ago

PLEASE HELP! I ONLY HAVE 10 MINUTES LEFT!

Mathematics

1 answer:

Simora [160]3 years ago

8 0

Answer:

Multiplacation

Step-by-step explanation:

It is the multiplacation roperty pbecauseyou are multipyling 10 times x and 3.

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_________are triangles with congruent angles and proportional sides.

Oksana_A [137]

Answer: Similar triangles

Step-by-step explanation:

They have congruent angles and similar sides

7 0

3 years ago

2 1/4 x 3 2/5<br> show all your work

monitta

Answer:

153/20

Step-by-step explanation:

I assume you're asking

$2 \frac{1}{4} \times 3 \frac{2}{5}$

$\frac{9}{4} \times \frac{17}{5}$

$\frac{153}{20}$

5 0

2 years ago

At the end of a party, 3/4 cup of dip is left. Jim divides 4/7 of the leftover dip equally between 2 friends. How much dip does

Vesnalui [34]

Answer:

First, we need to find the amount of dip that was divided amount the two friends.

We know that it is 4/7 of the left dip

therefore:

amount divided among the two friends = (4/7) x (3/4) = 3/7 of the original amount of dip.

This amount is divided among two friends,

therefore:

amount that each friend gets = (3/7) / (2) = 3/14 = 0.2 of the amount of dip

tell me if this is completely wrong because i'm not good at this

7 0

3 years ago

PLEASE ANSWER QUICKLY

Luba_88 [7]

29-11=18
18²+18²=p²
324+324=p²
p²=648
p=18√2
That's your answer.

6 0

3 years ago

Read 2 more answers

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