What is a counter example for, when it rains, it pours?

Cierra has 3 more than one half as many purses as Aisha. Define a variable and write an expression to represent the number of pu

SOVA2 [1]

Use P to represent the number of purses that Aisha has. 1/2P x 3 if P= 12 then 1/2 (12) x 3= 18

8 0

4 years ago

Convert C^w=212 into log form

dsp73

Answer:

C^w=212 into log form

$log_{c}(212) = w$

6 0

3 years ago

David correctly factored the expression m^2 ─ 12m ─ 64. Which expression did he write?

julia-pushkina [17]

Answer:

3

Step-by-step explanation:

because it's got factors of 64 which when u add gives -12

7 0

3 years ago

PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU B

kodGreya [7K]

Answer:

30 is the volume

Step-by-step explanation:

Multiple 6x5 to get your answer

6 0

3 years ago

Increasing numbers of businesses are offering child-care benefits for their workers. However, one union claims that more than 85

Lubov Fominskaja [6]

Answer:

The p value for this test is given $p_v = 0.1071$

Since the p value is higher than the significance level given of $\alpha=0.05$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workersis is not significantly higher than 0.85 or 85% at 5% of significance.

Step-by-step explanation:

We define the proportion of interest as p who represent the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workers

And we want to anaylze the following system of hypothesis:

Null hypothesis: $p \leq 0.85$

Alternative hypothesis: $p >0.85$

And the statistic for this test is given by:

$z =\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$

The p value can be calculated with this formula:

$p_v = P(z>z_{calc})$

And the p value for this test is given $p_v = 0.1071$

Since the p value is higher than the significance level given of $\alpha=0.05$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workersis is not significantly higher than 0.85 or 85% at 5% of significance.

4 0

3 years ago