True or False: 2.4 ÷ 0.6 = 0.04 x 6 =

Brian and Christian started keeping track of their workouts. Brian did 85 sit-ups first week and 90 sit-ups each week after that

dezoksy [38]

Answer:

Brian will have done 445, and Christian will have done 425.

6 0

3 years ago

Fast plz rkoxeooq be dvbdisksndv

Maksim231197 [3]

Answer:

○ $\displaystyle -\frac{9}{14}$

Step-by-step explanation:

$\displaystyle \frac{3}{4} \times -\frac{6}{7} = -\frac{18}{28} = -\frac{9}{14}$

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4 0

4 years ago

HELP!!!! CLICK ON THE PICTURE! THANK YOU

ArbitrLikvidat [17]

19 over 25
19/25= 0.76
there ya go

3 0

3 years ago

In his will, Mr. Banks left 50% of the value of his property to his

Rashid [163]

The value of Mr Banks's property was $250,000

What percentage of the property is left for charity?

The percentage of property left for charity can be determined as the 100% of the property minus all percentages given to others

% left for charity=100%-50%-22%-16%

% left for charity=12.00%

12.00% of the property=$30,000

1.00% of the property =$30,000/12

1.00% of the property =$2,500

100.00% of the property=$2500*100

100.00% of the property=$250,000

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6 0

2 years ago

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where

z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704± $2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}}$ ≈ -0.171 ±0.237 that is (−0.408, 0.066)

7 0

3 years ago