If seven cats killed seven rats in 7 min, how many cats are needed to kill 200rats in 100min

Who can give me this answer

-BARSIC- [3]

The believe the answer is 3/5

3 0

2 years ago

(FASTEST AND BEST ANSWER GETS BRAINLIEST)1

Semmy [17]

Answer:

Step-by-step explanation:

um

6 0

2 years ago

Colby is making a home video consisting of a 5-minute introduction followed by several short skits. Each skit is 8 minutes long.

Delvig [45]

Colby is making a home video consisting of a 5-minute introduction followed by several short skits. Each skit is 8 minutes long. If colby's video is 133 minutes long how many skits are in his video?

He would have 16 skits in his video.

3 0

3 years ago

Read 2 more answers

Please help !! 25 points !!!

EastWind [94]

Answer:

30

Step-by-step explanation:

70-40=30

8 0

3 years ago

One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe

sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who disagree with the statement = $\frac{1812}{2625}$ = 0.69

n = sample of respondents = 2625

p = population proportion of people who disagree with statement

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ , $0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ ]

= [0.667 , 0.713]

Therefore, 99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

5 0

3 years ago

If seven cats killed seven rats in 7 min, how many cats are needed to kill 200rats in 100min​

If seven cats killed seven rats in 7 min, how many cats are needed to kill 200rats in 100min