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2 years ago

What is 7/9 + 1/6. It's for my homework.

Mathematics

2 answers:

krek1111 [17]2 years ago

6 0

Answer:

17/18 as a fraction.

Art [367]2 years ago

4 0

Answer:

0.9444444

Step-by-step explanation:

you could just put 0.94

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Write y=5x-9 in standard form<br> Will give brainliest!

ArbitrLikvidat [17]

Answer:

5x - y = 9

Step-by-step explanation:

the equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

given

y = 5x - 9 ( subtract 5x from both sides )

- 5x + y = - 9 ( multiply through by - 1 )

5x - y = 9 ← in standard form

6 0

2 years ago

: The perimeter of a rectangular garden is 46. The length is 7 less than 4 times the width. What are the dimensions of the garde

olga55 [171]

Let the width be x, then length = 4x - 7
Perimeter = 2(length + width)
2(x + 4x - 7) = 46
2(5x - 7) = 46
10x - 14 = 46
10x = 60
x = 6

Therefore, width is 6 and length is 4(6) - 7 = 24 - 7 = 17

4 0

3 years ago

Rosie bought a puppy. She weighed it for 2 months. The first month it weighed 2 1/2 pounds. Then she weighed it again the next m

Dvinal [7]

The puppy weights 3.75 pounds now

3 0

3 years ago

Which polynomial represents the difference below? (7x^2+8)-(4x^2+x+6)

Alex787 [66]

I think the answer is :
3x^2 + 2 - x

i hope this helps :)

6 0

4 years ago

In a data distribution, the first quartile, the median and the means are 30.8, 48.5 and 42.0

kolbaska11 [484]

Answer:

$Q_3 = 56.45$ --- The third quartile

$Var = 370.18$ -- Variance

Step-by-step explanation:

Given

$Q_1 = 30.8$ -- First quartile

$Q_2 = 48.5$ --- Median

$\bar x = 42$ --- Mean

$Skp = -0.38$ --- Coefficient of skewness

Solving (9): The third quartile $Q_3$

This is calculated from

$Skp = \frac{Q_1 + Q_3 - 2Q_2}{Q_3 - Q_1}$

So, we have:

$-0.38 = \frac{30.8 + Q_3- 2*48.5}{Q_3 - 30.8}$

Cross Multiply

$-0.38 (Q_3 - 30.8)= 30.8 + Q_3- 2*48.5$

Open bracket

$-0.38Q_3 + 11.704= 30.8 + Q_3- 97.0$

Collect like terms

$-0.38Q_3 -Q_3= 30.8 - 97.0- 11.704$

$-1.38Q_3= -77.904$

Divide both sides -1.38

$Q_3 = \frac{-77.904}{-1.38}$

$Q_3 = 56.45$ --- approximated

Solving (b): The variance

First, calculate the standard deviation from:

$3IQR = 4SD$

$IQR= Q_3 - Q_1$

So:

$3IQR = 4SD$

$3(Q_3 - Q_1) = 4SD$

Make SD the subject

$SD = \frac{3}{4}(Q_3 - Q_1)$

$SD = \frac{3}{4}(56.45 - 30.8)$

$SD = \frac{3}{4}*25.65$

$SD = \frac{3*25.65}{4}$

$SD = \frac{76.95}{4}$

$SD = 19.24$

So, the variance is:

$Var = SD^2$

$Var = 19.24^2$

$Var = 370.18$

5 0

3 years ago