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3 years ago

I NEED HELP QUICKLY!!! What is the sum of the first 140 positive multiples of 3?

Mathematics

1 answer:

garri49 [273]3 years ago

5 0

Answer:

Step-by-step explanation:

The multiples of 3 are

3, 6, 9, 12, 15, ...........

The terms form an arithmetic sequence with common difference of 3.

The sum to n terms is

$S_{n}$ = $\frac{n}{2}$ [2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

Here a₁ = 3 and d = 3, thus

$S_{140}$ = $\frac{140}{2}$ [ (2 × 3) + (139 × 3) ]

= 70(6 + 417) ] = 70 × 423 = 29610 → C

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Mapoela walks 2 km 30 minutes. At that rate, how far could she walk in 1 ½ hours

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Answer:

Step-by-step explanation:

2=30 so 30= 1/2 or an hour so 2+2=4 witch makes an gour half of 2 is 1 so it would make 5

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2 years ago

An electrical power company is looking to expand into a new market. Before they commit to supplying the new area with electricit

TiliK225 [7]

Answer:

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Step-by-step explanation:

Given that an electrical power company is looking to expand into a new market. Before they commit to supplying the new area with electricity, they would like know the mean daily power usage for homes there.

Population std deviation = $\sigma = 50$

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4 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$