Ask question

Ask question

All categories

3 years ago

8

If student D can solve 60% of multiple choice questions and Student B can solve 51% of the multiple choice questions and Student

C can solve 57%. The students are unaware of each other.
a) At least one can be solved
b) Probability at most one of them can be solved
c) Probability that neither of them can be solved
d) Exactly two them can be solved

1 answer:

SOVA2 [1]3 years ago

3 0

Do you know answer or not please give me answer

You might be interested in

Help Me Please! I will give you 5 stars if your answer is correct!

Nadya [2.5K]

Answer:

N

Step-by-step explanation:

the absloute value of -4 is 4

when you take the absolute value, you change any negative number into a positive number by removing the negative sign

8 0

3 years ago

I need help please!!!

Jlenok [28]

Answer:

1. Explain what a medical sociologist means by "health and illness are social attributes influenced by social factors"

7 0

3 years ago

Helppp me plsssssssss<br><br>

Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42$
$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

$\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\$

$\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\$

${\large{\frak{\pmb{\underline{Additional\: information }}}}}$

MODE

Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

In a frequency distribution the class having maximum frequency is called the modal class.

${\bf{\underline{Formula\:for\: calculating\:mode:}}} \\$

${\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\$

Where,

$\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.$

$\small \blue{ \bigstar}$ $\sf \: f_{k}=frequency\:of\:the\:modal\:class$

$\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class$

$\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class$

$\small \purple{ \bigstar}$ $\sf \: h= width \:of\:the\:class\:interval$

7 0

3 years ago

A circle is centered at the point (-3, 2) and passes through the point (1, 5). The radius of the circle is ______ units. The poi

nalin [4]

Circle formula
(x-h)^2+(y-k)^2=r^2 where (h,k) is the center
and r=radius

to find the radius
we are given one of the points and the center
distnace from them is the radius
distance formula
D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$
points (-3,2) and (1,5)
D= $\sqrt{(1-(-3))^{2}+(5-2)^{2}}$
D= $\sqrt{(4)^{2}+(3)^{2}}$
D= $\sqrt{16+9}$
D= $\sqrt{25}$
D=5

center is -3,2
r=5
input
(x-(-3))^2+(y-2)^2=5^2
(x+3)^2+(y-2)^2=25 is equation
radius =5
input -7 for x and solve for y
(-7+3)^2+(y-2)^2=25
(-4)^2+(y-2)^2=25
16+(y-2)^2=25
minus 16
(y-2)^2=9
sqqrt
y-2=+/-3
add 2
y=2+/-3
y=5 or -1

the point (-7,5) and (7,-1) lie on this circle

radius=5 units
the points (-7,5) and (-7,1) lie on this circle

3 0

3 years ago

40.6 in numerical form

oksano4ka [1.4K]

What's your question so I could help?

6 0

3 years ago