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3 years ago

Please help me with this its’s for a test

Mathematics

1 answer:

Zinaida [17]3 years ago

8 0

The missing # is 3 & 6

(X+2)(x+3) = x^2+ 5x+6

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One third plus one half

LekaFEV [45]

<h2><em>Find the LCD of 1/3 and 1/2.</em></h2><h2><em>The LCD is 6.</em></h2><h2><em>1/3 in 6ths is 2/6.</em></h2><h2><em>1/2 in 6ths is 3/6.</em></h2><h2><em>2/6 + 3/6 = 5/6.</em></h2><h2><em>1/3 + 1/2 = 5/6.</em></h2><h2><em>Hope this helps and have a nice day.</em></h2>

6 0

3 years ago

Read 2 more answers

Will mark BRAINLIEST!

vazorg [7]

count them all together
put what you got as the denominator
and then put how many bracelets she made as the numerator

5 0

3 years ago

Help ineed to find this.

omeli [17]

She saved up a total of $82.60, and had to spend a total of $60.90, so she would have $21.70 left and need to save up another $180.40, hope this helps

5 0

3 years ago

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Mary can make a quilt in 3 days and Eva can make

Rina8888 [55]

Answer: in explanation

Step-by-step explanation

Sandi can perform 1/9 of the task in a day

Eva can perform 1/5 of the task in a day

Working together, they can perform

1/9 + 1/5 of the task per day

= 5/45 + 9/45 = 14/45 of the task in a day

or 45/14 days in the task = 3.2 days

It would take about 3.2 days to complete the task working togethe

7 0

3 years ago

Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

7 0

3 years ago