Ask question

Ask question

All categories

3 years ago

8

1] in a survey , one out of ten people named blue as there favorite color. tree out of five named red. if 1,00 people were inclu

ded in the survey .how many named blue or red as their favorite color ?
a] 330 people

b] 770 people

c] 440 people

d] 660 people

please help !!!!

1 answer:

GrogVix [38]3 years ago

4 0

It should be 70 because 60% plus 10% equals 70

You might be interested in

4/15 x 6 in simplest form

Alborosie

Answer:

1.6

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Xy′ = √(1 − y2 ), y(1) = 0

Ulleksa [173]

Answer:

The particular solution is $y=\sin (\ln|x|)$ .

Step-by-step explanation:

The given differential equation is

$xy'=\sqrt {1-y^2}$

It can be written as

$x\frac{dy}{dx}=\sqrt {1-y^2}$

Use variable separable method to solve the above equation.

$\frac{dy}{\sqrt {1-y^2}}=\frac{1}{x}dx$

Integrate both sides.

$\int \frac{dy}{\sqrt {1-y^2}}=\int \frac{1}{x}dx$

$\sin^{-1} y=\ln|x|+C$ .... (1)

It is given that y(1)=0. It means y=0 at x=1.

$\sin (0)=\ln|1|+C$

$0=0+C$

$0=C$

The value of constant is 0.

Substitute C=0 in equation (1) to find The required equation.

$\sin^{-1} y=\ln|x|+0$

Taking sin both sides.

$y=\sin (\ln|x|)$

Therefore the particular solution is $y=\sin (\ln|x|)$ .

7 0

3 years ago

1 and 2 i get them but 3-6 i don't get it

katovenus [111]

3) will be a horizontal line just forever stretching across the 7 y value never straying

4) will be a vertical line just forever stretching across the 1 x value never straying

5 & 6 are the ones I don't understand

5 0

3 years ago

Bree took a math test and scored 19 points, which was 95% of the total possible points on the test. What was the total number of

NISA [10]

Answer: 20 points hope this helps

3 0

2 years ago

Prove that every bounded increasing sequence is convergent

aleksklad [387]

The proof of this can be get with a slight modification. It can be prove that every bounded is convergent, If (an) is an increasing and bounded sequence, then limn → ∞an = sup{an:n∈N} and if (an) is a decreasing and bounded sequence, then limn→∞an = inf{an:n∈N}.

4 0

3 years ago