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2 years ago

Students received these scores: 70, 75, 80, 90, 95, 95, and 100. the score of 95 is what?

Mathematics

1 answer:

Ganezh [65]2 years ago

7 0

95 is the mode because there is the largest quantity of it

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Have a look of the drawing... What is "d" value? I have tried many methods but don't find how to solve... Pleaseeee

Phoenix [80]

I´d say "d" is the distance from the eye to the wall.
Now substracting 1.2-1 you´ll get the distance of the wall of the smallest triangle = 0.2 And you do 1.5-0.2= 0.3 that´s the distance of the wall of the other triangle. Then you solve everything with Pitagoras theorem. You have 2 rectangle triangles.
B+alfa=45°
tan^-1(0.2/d)=B
tan^-1(1.3/d)=alfa
THEN:
tan^-1(0.2/d)+tan^-1(1.3/d)=45°
Now you have 3 ecs and 3 variables.
alfa,B and "d"

3 0

3 years ago

Read 2 more answers

HELPP I‘LL GIVE BRAINLIEST!

TiliK225 [7]

Answer:

so Omar approximately receives 393€

please mark brainliest!♡♡

Step-by-step explanation:

2000/5.087= 393.15903

8 0

2 years ago

Javier invests $2,500 into an account 3.5% simple interest . What will be the total amount in the account after 5 years if no ot

katovenus [111]

Answer:d

Step-by-step explanation:

4 0

2 years ago

Solve. 2a + 3b = 5 6= a -5 a = 4 b = -1 1 a = 6 b=1 a=6 6 = -1 a=4 6 = 1<br>

slamgirl [31]

Answer:Solve. 2a + 3b = 5 6= a -5 a = 4 b = -1 1 a = 6 b=1 a=6 6 = -1 a=4 6 = 1

Step-by-step explanation:

3 0

2 years ago

In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

5 0

2 years ago