2 years ago

Please hurry it is 6th grade math b^{2}=\frac{4}{9}

Mathematics

1 answer:

allochka39001 [22]2 years ago

4 0

Can you type it in a easier way to read

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Evaluate-2.75 - f for f = -6.5

Stels [109]

Answer:

Step-by-step explanation:

we have f=-6.5

-2.75- (-6.5)= -2.75+6.5=3.75

4 0

2 years ago

2 dots plots with number lines going from 0 to 10. Plot A has 0 dots above 0, 1, and 2, 1 above 3, 2 above 4, 2 above 5, 2 above

alexandr402 [8]

Answer:

First one:

Both the mean and median are greater for Plot A than for Plot B

Step-by-step explanation:

Set A:

Mean:

[1×10 + 2×7 + 2×6 + 2×5 + 2×4 + 1×3]/10

= 5.7

Median:

Median position: (10+1)/2 = 5.5th value

(5+6)/2

Median = 5.5

Set B:

Mean:

[1×7 + 3×6 + 3×5 + 2×4 + 1×3]/10

= 5.1

Median:

Median position: (10+1)/2 = 5.5th value

(5+5)/2

Median = 5

Mean: A is greater

Median: A is greater

8 0

3 years ago

In Mr. Siegel’s class, 15% of the students are in the Scrapbooking Club and 45% are in the Cooking Club. His remaining students

Oksanka [162]

Answer:

Step-by-step explanation:

15% = 15/100= 0.15

45% = 45/100 = o.45

Remaining students: 40% = 40/100 = 0.4%

6 0

2 years ago

5. Complete the following equation using < >, or =<br> 7___24/4

Zepler [3.9K]

Answer:

7 > 24/4

Step-by-step explanation:

7 __ 24/4

simplify

7 __ 6

7 > 6

3 0

2 years ago

Read 2 more answers

Explain why S = {(5, 4, -2), (-15, -12, 6), (10, 8, -4)} is NOT a basis for R^3 (1 point Sis linearly dependent and spans R3 S i

earnstyle [38]

$S$ would be a basis for $\mathbb R^3$ if

(1) the vectors in $S$ are independent, and

(2) the vectors span $\mathbb R^3$ .

Linear independence requires that $c_1=c_2=c_3=0$ is the only solution to

$c_1(5,4,-2)+c_2(-15,-12,6)+c_3(10,8,-4)=(0,0,0)$

These vectors are not linearly independent because if $c_1=3$ , $c_2=1$ , and $c_3=0$ , we have

$3(5,4,-2)+(-15,-12,6)=(15-15,12-12,-6+6)=(0,0,0)$

so $S$ is not a basis for $\mathbb R^3$ .

7 0

3 years ago