What is the quotient? −4 1/2÷(−2 2/3)

irina1246 [14]

Answer:

B. Dog 1's position is Dog 2's position reflected across the x-axis; only the signs of the y-coordinates of Dog 1 and Dog 2 are different.

Step-by-step explanation:

First it is reflected on the x-axis because it is above and below. if its was the y-axis (which it is not) it would be side to side, if that makes sense.

Secondly the x values are the same, they are both -6. So, the y values are the ones that are different.

5 0

3 years ago

How do I solve 18-4K=-2 step by step

jek_recluse [69]

Answer:5

Step-by-step explanation: 18-4K = -2

-18 -18

-4K = -20

-4 -4

K=5

3 0

4 years ago

Use properties of exponents to complete the equivalent expression. (Take a look at the image).

kondaur [170]

The exponent would b 15

U just gotta solve it than u get 3375 than look for the square root of that and it’s 15over15

6 0

3 years ago

The desired percentage of Silicon Dioxide (SiO2) in a certain type of aluminous cement is 5.5. To test whether the true average

Reil [10]

Answer:

a) Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

b) The test statistic

|t| = |-3.33| = 3.33

c) P - value lies between in these intervals

0.001 < P < 0.005

Step-by-step explanation:

Step( i ):-

Given data the Population mean 'μ' = 5.5

The small sample size 'n' = 16

The sample mean (x⁻) = 5.25

Given the percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.

Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

Level of significance ∝ = 0.01

Step(ii):-

The test statistic

$t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }$

On calculation , we get

t = -3.33

|t| = |-3.33| = 3.33

Step(iii):-

P - value

The degrees of freedom γ = n-1 = 16-1 =15

The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005

0.001 < P < 0.005

Condition(i)

P - value < ∝ then reject H₀

Condition(ii)

P - value > ∝ then Accept H₀

we observe that 0.001 < P < 0.005

P- value < 0.01

we rejected H₀

(or)

The tabulated value = 2.60 at 0.01 level of significance with '15' degrees of freedom

The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom

The null hypothesis is rejected

Conclusion:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5

4 0

3 years ago

Help! Hurry!!! NO SPAM!!!!!!!! I will mark brainliest if you get it correct, and show your work. Determine whether f is an expon

Roman55 [17]

F it’s not an appitional

5 0

3 years ago