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3 years ago

Which decimal is equivalent to 1 5/8?

Mathematics

1 answer:

mojhsa [17]3 years ago

3 0

Answer: 1 5/8 is equivalent to 1.625

Step-by-step explanation:

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What is the answer to 1/4+1/2

Bas_tet [7]

Answer:

0.75

Step-by-step explanation:

7 0

3 years ago

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Martiza remembers her pin because it is between 1000 and 1500 and it is a product of two consecutive prime numbers what is her p

Anna71 [15]

Answer:

Her pin must be 1147

Step-by-step explanation:

A little trick I have up my sleeve when I have these kind of questions is that I find the square root of each number, then find what is prime.

$\sqrt{1000} = 31.6$

$\sqrt{1500} = 38.7$

Since the only prime numbers in the range of 31 and 38 is 31 and 37.

We can then multiply 31 by 37 to get 1147.

6 0

3 years ago

Use a calculator to find the approximate value of the expression. Round the answer to two decimal places. The cosine of the comp

taurus [48]

Answer:

B) 0.85

Step-by-step explanation:

By definition, complementary angles add to 90 degrees.

The given angle 58 degrees is complementary to 90-58 = 32 degrees. The two angles 58 and 32 add to 90.

Therefore, cos(32) = 0.848 = 0.85

4 0

3 years ago

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£245 £140 £525 £174 £195 £163 £140 what is the mode?

Blababa [14]

The mode is the number that repeats itself the most in a sequence. In this case, the mode will be £140 since it repeats itslef the most (2 times).

Hope it helped,

BioTeacher101

5 0

3 years ago

Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1)

ioda

(1) Looks like the joint density is

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0$

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

$\displaystyle\int_0^4\int_y^4 cxy\,\mathrm dx\,\mathrm dy=\int_0^4\frac{cy}2(4^2-y^2)=32c=1$

$\implies\boxed{c=\dfrac1{32}}$

(2) The region in which X > 2 and Y < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

$P(X>2,Y$

(3) Are you supposed to find the marginal density of X, or the conditional density of X given Y?

In the first case, you simply integrate the joint density with respect to y:

$f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^x\frac{xy}{32}\,\mathrm dy=\begin{cases}\frac{x^3}{64}&\text{for }0$

In the second case, we instead first find the marginal density of Y:

$f_Y(y)=\displaystyle\int_y^4\frac{xy}{32}\,\mathrm dx=\begin{cases}\frac{16y-y^3}{64}&\text{for }0$

Then use the marginal density to compute the conditional density of X given Y:

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}\frac{2xy}{16y-y^3}&\text{for }y$

6 0

3 years ago