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Vladimir79 [104]

3 years ago

9

In a hostel, 120 students had provisions of food for 60 days. After 20 days, some students were added to the hostel, the provisi

on of food finished 10 days ago earlier. How many students were added?

1 answer:

DochEvi [55]3 years ago

8 0

Answer:

If I'm right, the answers should be 40.

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Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalat

ella [17]

Step-by-step explanation:

n = 52

x bar = 122.31

standard deviation sd = 38.75

1. the hypothesis:

null hypothesis

h0: μ ≤ 35

alternative hypothesis

h1: μ > 35

2.

we have been given the sd of the sample but not that of the population. so what we are supposed to use here is the t test and not the z test. the following conditions have to be met.

population has to be normal
sample size has to be more than 30. we have sample size = 52 in this question.

3.

= $\frac{122.31-35}{38.75/\sqrt{52} }$

= 87.31/5.37

= 16.26

using the T distribution function in excel, the p value was calculated and found to be approximately equal to 0.

TDIST(16.26, 51, 1)

since p value is very small we reject the null and accept the alternate hypothesis.

4. from the result above the answer to this question is yes

8 0

3 years ago

-35÷6= <br><br> -350÷6-350-6= <br><br> 350 - 60 =

eduard

Answer:

The first one is 5.83 or -5 5/6

The second one is -414 1/3

The last one is 290

4 0

2 years ago

Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3

NikAS [45]

Answer:

(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

1. $f(x)=-2x+1$

If we compare this equation with the slope intercept form of a line which is given as

$y=mx+c$

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

$y=-2(1)+1$

$y=-2+1 = -1$

Let us find another point by putting x = 2 and solving it for y

$y=-2(2)+1$

$y=-4+1 = -3$

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

$y=-2(-2)+1$

$y=+1 = 5$

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) , (2,-3) and (-2,5) we joint them on line to obtain our line

2.

$g(x)=y=x^2-2x-3$

$y=x^2-2x+1-1-3$

$y=(x-1)^2-4$

$(y+4)=(x-1)^2$

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; $y=y=(0)^2-2(0)-3=0-0-3=-3$ ; (0,-3)

ii)x=-1 ; $y=(-1)^2-2(-1)-3=1+2-3=0$ ; (-1,0)

iii) x=2 ; $y=(2)^2-2(2)-3 = 4-4-3 =-3$ ;(2,-3)

iii) x=1 ; $y=(1)^2-2(1)-3 = 1-2-3 =-4$ ;(1,-4)

iii) x=-2 ; $y=(-2)^2-2(-2)-3 = 4+4-3 =5$ ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.

3 0

4 years ago

Prove that 1+cosA/sinA + sinA/1+cosA=2cosecA

Alla [95]

$\frac{1+cos\alpha}{sin\alpha}+\frac{sin\alpha}{1+cos\alpha}=2cosec\alpha\\\\L=\frac{(1+cos\alpha)(1+cos\alpha)+sin\alpha\cdot sin\alpha}{sin\alpha(1+cos\alpha)}=\frac{1+2cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha(1+cos\alpha)}\\\\=\frac{1+2cos\alpha+1}{sin\alpha(1+cos\alpha)}=\frac{2+2cos\alpha}{sin\alpha(1+cos\alpha)}=\frac{2(1+cos\alpha)}{sin\alpha(1+cos\alpha)}\\\\=\frac{2}{sin\alpha}=2\cdot\frac{1}{sin\alpha}=2cosec\alpha=R$

5 0

3 years ago

The first two steps in determining the solution set of the system of equations y=x to the second power -2x-3 and y=-x+3

Arisa [49]

The solution set for the both equation is (3,0) and (-1,0)

We have given that the equation

$y=x^2-2x-3 \ and\ y=x+3$

We have to determine the solution to the first equation

<h3>What is the quadratic equation formula?</h3>

$\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-3\right)}}{2\cdot \:1}$

$x_{1,\:2}=\frac{-\left(-2\right)\pm \:4}{2\cdot \:1}$

$x_1=\frac{-\left(-2\right)+4}{2\cdot \:1},\:x_2=\frac{-\left(-2\right)-4}{2\cdot \:1}$

$x=3,\:x=-1$

Now use this value we have to find the value of y

When x=3 then

y=(3)^2-2(3)-3

y=9-6-3

y=0

When x=-1 then y=1-2(-1)-3

y=1+2-3

y=0

Therefore the solution set for both equations is (3,0) and (-1,0).

To learn more about the solution set visit:

brainly.com/question/10588366

#SPJ1

7 0

2 years ago