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3 years ago

How many bits are needed to represent the decimal number 200?

1 answer:

7 0

Log 2 200 = In 200 ã• In 2 = 7.6... â need 8 bits. If a signed number is being stored, then 9 bits would be needed as one as one would be needed to indicate the sign of the number.

8 (assuming unsigned numbers - i.e., you don’t reserve a bit for the sign.

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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

It’s less than 60 and greater than 55. If you add the digits the sum is 13.

labwork [276]

Between 60 and 55, 5+8=13

6 0

3 years ago

Find the next 3 terms in the geometric sequence -36, 6, -1, 1/6, ...

timama [110]

we are given

geometric sequence -36, 6, -1, 1/6, ...

first term is -36

$a_1=-36$

now, we can find common ratio

$r=\frac{6}{-36}$

$r=-\frac{1}{6}$

now, we can find nth term

$a_n=a_1(r)^{n-1}$

now, we can plug values

and we get

$a_n=36(-\frac{1}{6})^{n-1}$

now, we can find 5th term , 6th term, 7th term

fifth term:

$a_5=36(-\frac{1}{6})^{4}$

$a_5=\frac{1}{36}$

sixth term:

$a_6=36(-\frac{1}{6})^{5}$

$a_6=-\frac{1}{216}$

seventh term:

$a_7=36(-\frac{1}{6})^{6}$

$a_7=\frac{1}{1296}$

so, next terms are

$a_5=\frac{1}{36}$ , $a_6=-\frac{1}{216}$

, $a_7=\frac{1}{1296}$ .............Answer

6 0

3 years ago

Ophelia is making homemade spaghetti sauce by combining 48 oz of tomato paste with 6 cups of water.how many cups of water are ne

kobusy [5.1K]

1 cup per 8 oz

48/6 = 8

every 1 cup of water 8 oz of tomato paste.

3 0

3 years ago

Read 2 more answers

Find the two intersection points

bogdanovich [222]

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$