All categories

3 years ago

2r - 3.1 =1.7 ; 4t + 3.5 =12.5 ; 8m - 5.5 =10.1 i really need help plzzzzzzz

1 answer:

6 0

$2r - 3,1 = 1,7
$
Keep the numbers with incognito on the left and the numbers without incognito on the right. Exchanging the signs when changing sides if applicable.
 $2r = 1,7 + 3,1$
$2r = 4,8$
$r = \frac{4,8}{2}$
$\boxed{r = 2,4}$

--------------------------------
$4t + 3,5 = 12,5$
Keep the numbers with incognito on the left and the numbers without incognito on the right. Exchanging the signs when changing sides if applicable.
 $4t = 12,5-3,5$
$4t = 9,0$
$t = \frac{9,0}{4}$
$\boxed{t = 2,25}$
-------------------------------------

$8m - 5.5 = 10.1$
Keep the numbers with incognito on the left and the numbers without incognito on the right. Exchanging the signs when changing sides if applicable.
 $8m = 10,1 + 5,5$
$8m = 15,6$
$m = \frac{15,6}{8}$
$\boxed{m = 1,95}$

You might be interested in

Based on the Nielsen ratings, the local CBS affiliate claims its 11:00 PM newscast reaches 41 % of the viewing audience in the a

ZanzabumX [31]

Answer:

1) Null hypothesis: $p\geq 0.41$

Alternative hypothesis: $p < 0.41$

2) $\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

3) $z_{crit}=-2.33$

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

4) $z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017$

5) $z_{crit}=-1.28$

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

6) We see that $|t_{calculated}|$ so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

7) Null hypothesis: $p\geq 0.41$

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X represent the people indicated that they watch the late evening news on this local CBS station

$\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

$p_o=0.41$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) Part 1We need to conduct a hypothesis in order to test the claim that 11:00 PM newscast reaches 41 % of the viewing audience in the area: Null hypothesis:[tex]p\geq 0.41$

Alternative hypothesis: $p < 0.41$

Part 2

$\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

Part 3

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.01 of the area on the left and on this case this value is :

$z_{crit}=-2.33$

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

Part 4

When we conduct a proportion test we need to use the z statisitc, and the is given by:

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

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017$

Part 5

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.1 of the area on the left and on this case this value is :

$z_{crit}=-1.28$

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

Part 6

We see that $|t_{calculated}|$ so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

Part 7

Null hypothesis: $p\geq 0.41$

4 0

3 years ago

You are saving to buy a new phone you have $150 in your bank account and can save $55 each week assuming you don’t spend any mon

FromTheMoon [43]

Answer:

150+55w