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3 years ago

3x + 2<5 ? Show work please

Mathematics

1 answer:

Ganezh [65]3 years ago

7 0

Answer:

x < 1

Step-by-step explanation:

3x + 2<5 (subtract 2 on each side)

3x< 3 (Divide 3 on both sides)

x < 1

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1/2 x 2/4 x 3/6 x 4/8 x 5/10 x 6/12 simplified

Talja [164]

Answer: 3

Step-by-step explanation: Because, 1/2=2/4=3/6=4/8=5/10=6/12 are all the same. You can test is out by simply factoring out the biggest factor on both sides. For example, 2/4 can be simplified to 1/2, because 2/4=(2/2)/(4/2) you can divide both sides by two. Hope this helps, <3!

8 0

3 years ago

Avicenna, a major insurance company, offers five-year life insurance policies to -65 year-olds. If the holder of one of these po

Alecsey [184]

Answer:

The question is about the least amount to charge each policyholder as premium

The least premium is $484

Step-by-step explanation:

The least amount of premium to charge for this policy is the sum of the expected values of outcome of both instances of policyholder dying before the age of 70 and living after the age of 70 years

expected value of dying before 70 years=payout*probability=$24,200*2%=$484

Expected of living after 70=payout*probability=$0*98%=$0

sum of expected values=$484+$0=$484

Note that payout is nil if policyholder lives beyond 70 years

The premium of $573 means that a profit of $89 is recorded

5 0

3 years ago

MATH QUESTION, PLEASE HELP !!!<br> (will mark branliest )

Aleksandr [31]

Answer:

I am not certain if its right

Step-by-step explanation:

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3 years ago

I need help ok help meeee

uysha [10]

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3 years ago

A genetics experiment involves a population of fruit flies consisting of 1 male named Bart and 3 females named Charlene, Diana,

kenny6666 [7]

Answer:

(b) $\frac{1}{2}$

Step-by-step explanation:

1 male named Bart (b)

3 females named Charlene(c), Diana(d), and Erin(e).

Since there is replacement, the possible samples are:

bb, bc, be, bd, cb, cc, cd, ce, db, dc, dd, de, eb, ec, ed, and ee.

Total Number of pairs = 16

Event of picking 2 males:bb

Event of 1 male:bc,cb,bd,db,be,eb

Event of picking 0 males:

cc, cd, ce, dc, dd, de, ec, ed, and ee.

$\begin{equation*} \begin{matrix}Proportion of males & Probability \\0 & 9/16 \\1 & 6/16 \\2 & 1/16 \end{matrix} \end{equation*}$

b. The mean of the sampling distribution

$\mu = (\frac{9}{16} X0 ) +(\frac{6}{16} X 1)+(\frac{1}{16} X 2) = \frac{8}{16} = \frac{1}{2}$

c. No; the proportion of males is $\frac{1}{4}$ while the mean is $\frac{1}{2}$ .

B. No, the sample mean is not equal to the population proportion of males. These values are not always equal, because proportion is an unbiased estimator.

7 0

3 years ago