Please help on at least 2 of these!! I'll give brainliest!!

3a−15=4a−3a I NEED HELP WITH THIS ASAP PLZ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

GaryK [48]

Answer:

a = 15/2

Step-by-step explanation:

3a−15=4a−3a

Combine like terms

3a - 15 = a

Subtract 3a from each side

3a-3a−15=a−3a

-15 = -2a

Divide each side by -2

-15/-2 = -2a/-2

15/2 =a

8 0

3 years ago

Read 2 more answers

Find the common ratio of the geometric sequence –5,–30, -180,...

Murljashka [212]

Answer:

6

Step-by-step explanation:

You have to find what to multiply by to get to the next term

-30 ÷ -5 = 6

- 180 ÷ -30 = 6

therfore the common ratio is 6

5 0

3 years ago

Read 2 more answers

What is 2 13/56 simplified?

Paladinen [302]

Answer:

It already is.

Step-by-step explanation:

13/56 is already in the simplest form. It can be written as 0.232143 in decimal form (rounded to 6 decimal places).

What is 125/56 simplified? 125/56 as a fraction in simplest form is 125/56. Here we will show you how to simplify, reduce fraction 125/56 in its lowest ...

3 0

2 years ago

A manufacturing process produces semiconductor chips with a known failure rate of . If a random sample of chips is selected, app

AleksandrR [38]

Answer:

The probability that at least 14 of the chips will be defective is 0.6664.

Step-by-step explanation:

The complete question is:

A manufacturing process produces semiconductor chips with a known failure rate of 5.4%. If a random sample of 300 chips is selected, approximate the probability that at least 14 will be defective. Use the normal approximation to the binomial with a correction for continuity .

Solution:

Let X = number of defective chips.

The probability that a chip is defective is, p = 0.054.

A random sample of n = 300 chips is selected.

A chip is defective or not is independent of the other chips.

The random variable X follows a Binomial distribution with parameters n = 300 and p = 0.054.

But the sample selected is too large.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np=300\times 0.054=16.2>10\\n(1-p)=300\times (1-0.054)=283.8>10$

Thus, a Normal approximation to binomial can be applied.

So, $X\sim N(\mu =16.2,\ \sigma^{2}=15.3252)$ .

Compute the probability that at least 14 of the 300 chips will be defective as follows:

Use continuity correction:

P (X ≥ 14) = P (X > 14 + 0.50)

= P (X > 14.50)

$=P(\frac{X-\mu}{\sigma}>\frac{14.50-16.20}{\sqrt{15.3252}})$

$=P(Z>-0.43)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that at least 14 of the chips will be defective is 0.6664.

5 0

3 years ago

Find the indicated nth term show your solution 8,15,22,29,...,a18

Damm [24]

Answer:

127

Step-by-step explanation:

Hey there! :)

Let us find the 18th term — first, there are two main types of sequences.

Arithmetic Sequence — a sequence that has same difference or common difference.
Geometric Sequence — a sequence that has same ratio.

But if you notice, it cannot be geometric sequence because if we do ratio test, there are no common ratios.

Hence, it can only be arithmetic sequence.

Common Difference

$\displaystyle \large{a_{n + 1} - a_n = d}$

d is our common difference.

a_n+1 is the next term of a_n.

Let's check the common difference!

15-8 = 7

22-15 = 7

29-22 = 7

Therefore, 7 is our common difference.

General Terms of Arithmetic Sequence

$\displaystyle \large{a_n = a_1 + (n - 1)d}$

We know that a1 is the first term of sequence which is 8.

Our common difference is 7.

We want to find the 18th term — therefore,

$\displaystyle \large{a_{18} = 8 + (18 - 1)7} \\ \displaystyle \large{a_{18} = 8 + (17)7} \\ \displaystyle \large{a_{18} = 8 + 119} \\ \displaystyle \large{a_{18} = 127}$

Therefore, the 18th term is 127.

4 0

3 years ago