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3 years ago

Complete the chart please :)

Mathematics

2 answers:

Viktor [21]3 years ago

8 0

Answer:

x=9 output 1=20.25

Output 2 equals 27

Step-by-step explanation:

This function is constant, so you do input times 2.25 for each.

The box below 6 is 9 because we are counting by 3

9 x 2.25=20.25

12 x 2.25=27

Hope this helped :)

aleksley [76]3 years ago

4 0

Answer:

Step-by-step explanation:

muffin times 2.25=output

so 12x2.25=27

and also x=9

so 9x2.25=20.25

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N is the sum of three consecutive primes and is also the product of two 2-digit primes, what is the least possible value of N

Nookie1986 [14]

Hello from MrBillDoesMath!

Answer: N = 143

Discussion:

This one took some trial and error! At first I listed all 2 digit primes, looked at the list, but didn't know how to proceed. So, I took the smallest 2 digit primes numbers: 11 and 13 and wondered if their product, 13*11 = 143, could be represented as the sum of 3 consecutive primes. I went back to my list of primes, added groups of three consecutive numbers that seemed to be in the right range to give the desired sum, and stumbled on 43, 47, and 53!

43 + 47 + 53 = 143 !

Therefore N = 143. It's the sum of 43, 47, and 53 as well as the product of 11 and 13.

Thank you,

MrB

7 0

3 years ago

At a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective pa

11Alexandr11 [23.1K]

Answer:

Probability that fewer than 2 of these parts are defective is 0.604.

Step-by-step explanation:

We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.

A random sample of 7 parts produced by this machine is chosen.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 7 parts

r = number of success = fewer than 2

p = probability of success which in our question is % of defective

parts produced by one of the machine, i.e; 19%

LET X = Number of parts that are defective

So, it means X ~ Binom(n = 7, p = 0.19)

Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)

P(X < 2) = P(X = 0) + P(X = 1)

= $\binom{7}{0}\times 0.19^{0} \times (1-0.19)^{7-0}+ \binom{7}{1}\times 0.19^{1} \times (1-0.19)^{7-1}$

= $1 \times 1 \times 0.81^{7} +7 \times 01.9^{1} \times 0.81^{6}$

= 0.604

Therefore, the probability that fewer than 2 of these parts are defective is 0.604.

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3 years ago

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4 years ago

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Aleksandr-060686 [28]

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3 years ago

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V125BC [204]

Answer:

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3 years ago