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3 years ago

I need help with a problem

Mathematics

2 answers:

mash [69]3 years ago

5 0

Answer:

i dont see an image

Step-by-step explanation:

matrenka [14]3 years ago

4 0

I got chu what is it

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9 meters higher than altitude X

aleksklad [387]

X+9 ? I'm not sure if that was the kind of answer you were looking for but that's what I got out of that.

7 0

3 years ago

Assume that different groups of couples use the XSORT method of gender selection and each couple gives birth to one baby. The XS

olganol [36]

Answer:

Mean and Standard deviation for the numbers of girls in groups of 36 births are 18 and 3 respectively.

Step-by-step explanation:

We are given that he X SORT method is designed to increase the likelihood that a baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5.

Now consider a group consisting of 36 couples.

The above situation can be represented through binomial distribution;

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

where, n = number of trials (samples) taken = 36 couples

r = number of success

p = probability of success which in our question is probability

of a girl, i.e.; p = 0.5

Let X = Numbers of girls in groups of 36 births

So, X ~ Binom(n = 36, p = 0.5)

Now, mean for the numbers of girls in groups of 36 births is given by;

Mean, E(X) = $n \times p$ = $36 \times 0.5$ = 18

Also, standard deviation for the numbers of girls in groups of 36 births is given by;

Standard deviation, S.D.(X) = $\sqrt{n\times p\times (1-p)}$

= $\sqrt{36\times 0.5\times (1-0.5)}$

= 3

3 0

3 years ago

6. If the net investment function is given by

Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated capital for the period.

(a) The capital formation from the end of the second year to the end of the fifth year is approximately 298.87.

(b) The number of years before the capital stock exceeds $100,000 is approximately 46.15 years.

Reasons:

(a) The given investment function is presented as follows;

$I(t) = 100 \cdot e^{0.1 \cdot t}$

(a) The capital formation is given as follows;

$\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot e^{0.1 \cdot t}} + C$

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore, for the three years; Year 3, year 4, and year 5, we have;

$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

$\displaystyle \mathbf{1000 \cdot e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000$

$\displaystyle e^{0.1 \cdot t}} = 101$

$\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15$

The number of years before the capital stock exceeds $100,000 ≈ 46.15 years.

Learn more investment function here:

brainly.com/question/25300925

6 0

2 years ago

7b Evaluate when b= -1 b-3

tiny-mole [99]

Answer:

b-3

-1-3

-4

is the correct answer

5 0

3 years ago

Please help and show work!! Even numbers ONLY!! Will give brainliest!!

Aliun [14]

You can do like this
This is my answer 1 to 10
Do it now

5 0

3 years ago

I need help with a problem​

I need help with a problem