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3 years ago

In a certain arithmetic sequence, if -4 is the first term and 10 is the third term, which term is 157?

Mathematics

2 answers:

MAXImum [283]3 years ago

5 0

Answer:

157 is the 24th term

Step-by-step explanation:

The formula for arithmetic sequences is an=a1+(n-1)d,

where an is the nth term, a1 is the first term, n is the term position, and d is the common difference

First, we need to find the common difference using the 10.

Since it's the third term, we have 10=-4+(3-1)d

Solving for d gives us d=7

Next, we need to find the term number of 157

Thus, we have 157=-4+(n-1)*7

Solving for n gives us n=24

We can check it like this: Following order of operations (i.e. PEMDAS) gives us 24-1=23*7=161-4=157

likoan [24]3 years ago

4 0

Answer:

a₁₅₇ = 1,088

Step-by-step explanation:

aₙ = a₁ + d(n-1); where 'd' if common difference

a₁₅₇ = -4 + 7(156)

= -4 + 1092

= 1,088

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A,D,E IS THE RATIOS EUQIVALENT TO8:5

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This is how I'd solve it:

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Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$

and the probability both do not work is:

$=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833$

Hence, Probability that atleast does not work is:

0.5+0.0833=0.5833

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zhenek [66]

The answer is d!!! Hope that helps!!

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