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skad [1K]
3 years ago
9

The box plot represents the number of minutes it takes for the students in a class to run a mile. A number line goes from 0 to 1

4. The whiskers range from 5 to 13, and the box ranges from 7 to 9. A line divides the box at 8. What is the minimum of the data? 0 5 7 13
Mathematics
2 answers:
Stels [109]3 years ago
8 0

Answer:

b

Step-by-step explanation:

alina1380 [7]3 years ago
5 0

Answer:

B.5

Step-by-step explanation:

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It is 3{:}32\text{ p.M.}3:32 p.M.3, colon, 32, start text, space, p, point, m, point, end text and Emma has gymnastics lessons a
Verizon [17]

Answer:

148 minutes

Step-by-step explanation:

It is 3{:}32\text{ p.M.}3:32 p.M.3, colon, 32, start text, space, p, point, m, point, end text and Emma has gymnastics lessons at 6{:}00\text{ p.M.}6:00 p.M.6, colon, 00, start text, space, p, point, m, point, end text How many minutes are there until Emma's gymnastics lessons?\

The time is 3:32 pm

Emma's gymnastics lesson is at 6:00 pm

6:00 pm - 3:32 pm = 2 hours 28 minutes

That is, we have 28 minutes before 4:00 pm

And 2 hours from 4:00 pm to 6:00 pm making a total of 2 hours 28 minutes

How many minutes are there until Emma's gymnastics lessons?

Convert 2 hours 28 minutes to minutes

1 hour = 60 minutes

2 hours = 60 minutes × 2 = 120 minutes

120 minutes + 28 minutes = 148 minutes

2 hours 28 minutes to minutes = 148 minutes

4 0
2 years ago
Which figure must be a rhombus?
Anna71 [15]
The answer is letter choice B
7 0
3 years ago
Read 2 more answers
Please help me, i promise its worth it!!!
Rashid [163]

\\ \sf\longmapsto 2(L+B)=55

\\ \sf\longmapsto 2(\dfrac{4}{3}x+x)=55

\\ \sf\longmapsto \dfrac{8}{3}x+2x=55

\\ \sf\longmapsto \dfrac{8x+6x}{3}=55

\\ \sf\longmapsto \dfrac{14x}{3}=55

\\ \sf\longmapsto 14x=165

\\ \sf\longmapsto x=11.78

\\ \sf\longmapsto x\approx 12

Now

  • B=x=12
  • L=4/3(12)=4(4)=16
5 0
2 years ago
Read 2 more answers
Which expression is the same as 10÷4?
prisoha [69]
10/4 is not the simplified version that it can be. So 5/2 would be the same expression.
3 0
2 years ago
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Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
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