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3 years ago

11

Miguel reads 3/4 chapters in 4/5 hours. What is the unit rate in chapters per hour?

1 answer:

kap26 [50]3 years ago

8 0

If its multiplication sipmlest form is, 6/10
If its addition simplest form is, 7/9
If its any other just tell me! Hope i was right :p

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The absolute value of a negative fraction is always greater than 1, true of false?!

Fantom [35]

Answer: False

Step-by-step explanation: False because zero is negative or positive. The absolute value of any number could also include the absolute value of 0, which would be 0. Thus, the absolute value of any rational number is not always greater than zero, it can be zero as well. However, it is true that the absolute value of any rational number can never be negative.

Rational Number definition: Rationals contain whole numbers, integers, decimals, fractions, basically most numbers or any numbers.

6 0

2 years ago

Non proportional or proportional ?<br> y= 3.75x + 2

Serjik [45]

Answer- non proportional

7 0

3 years ago

Read 2 more answers

The high temperature is forecast to be 5.4 F . What is this temperature in degrees Celsius ()?

AleksAgata [21]

Answer:

-14c

Step-by-step explanation:

I’m using the actual values 1F = -17C so just times that by 5.4

7 0

3 years ago

In an effort to determine the most effective way to teach safety principles to a group of employees, four different methods were

musickatia [10]

Answer:

critical value = 5.29

Step-by-step explanation:

Given that they are divided into 4 groups and a sample of 5 test was selected

N = 5 * 4 = 20

k = 4

∝ = 0.01

Df for numerator ( SS group )= k - 1 = 3

Df for denominator ( SSE group ) = N - k = 20 - 4 = 16

DF ( degree of freedom )

Next we will use the F table to determine the critical value

Critical value = $F_{16,3,0.01}$ = 5.29

5 0

3 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

1.

$m^{-1}=\dfrac{1}{m}$

2.

$q^0=1$

3.

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

4.

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

5.

$m^{-1}=\dfrac{1}{m}$

6.

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

7.

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

8.

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago