The data shows the weights of 2 groups of house cats in pounds.

3. PLEASE HELP! Jenna just opened a checking account at the bank. Within the first month, she deposited three checks for $34.98,

sashaice [31]

Answer:C) She would have $216.88.

Step-by-step explanation:

A) Deposits: +34.98; +51.02; +51.22

Withdrawals: -5.23; -5.22; -7.89

B) -7.89, -5.23, -5.22, +34.98, +51.02, +51.22

The smallest amount will be the largest withdrawal, as it takes away from the account; the largest amount will be the largest deposit, as it adds to the account.

98 + 34.98 = 132.98 + 51.02 = 184.00 + 51.22 = 235.22 - 7.89 = 227.33 - 5.23 = 222.10 - 5.22 = 216.88

8 0

3 years ago

Read 2 more answers

The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to d

k0ka [10]

Part A:

From the central limit theorem, since the number of samples is large enough (up to 30), the mean of the the mean of the average number of moths in 30 traps is 0.6.

Part B:

The standard deviation is given by the population deviation divided by the square root of the sample size.

$standard \ deviation= \frac{\sigma}{\sqrt{n}} = \frac{0.4}{\sqrt{30}} = \frac{0.4}{5.477} =0.073$

Part C:

The probability that an approximately normally distributed data with a mean, μ, and the standard deviation, σ, with a sample size of n is greater than a number, x, given by

$P(X\ \textgreater \ x)=1-P(X\ \textless \ x) \\ \\ =1-P\left(z\ \textless \ \frac{x-\mu}{\sigma/\sqrt{n}} \right)$

Thus, given that the mean is 0.6 and the standard deviation is 0.4, the probability that the average number of moths in 30 traps is greater than 0.7 given by:

$P(X\ \textgreater \ 0.7)=1-P(X\ \textless \ 0.7) \\ \\ =1-P\left(z\ \textless \ \frac{0.7-0.6}{0.073} \right)=1-P(z\ \textless \ 1.369) \\ \\ =1-0.91455=\bold{0.0855}$

7 0

4 years ago

The distance from the center of a round table top to the edge of the table top is 4 ft. What is the area of. the table top

vladimir2022 [97]

I am not sure on this one but I am sure someone will answer soon! :)

7 0

4 years ago

Please Help, Thanks! (Need them quick) 29. 1 to 5

SCORPION-xisa [38]

You'are welcome :))))))))))

7 0

3 years ago

Suppose that the distribution is bell-shaped. If approximately 99.7% of the lifetimes lie between 568 hours and 1066 hours, then

irakobra [83]

Answer:

$\sigma =\frac{478}{6}=79.667$

Step-by-step explanation:

The empirical rule, also referred to as "the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)". The empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

And on this case since we are within 3 deviations (because we have 99.7% of the data between 568 and 1066hours), the result obtained using the z score agrees with the empirical rule.

So on this case we can find the standard deviation on this ways:

$\mu -3\sigma = 568$ (1)

$\mu +3\sigma = 1066$ (2)

If we subtract conditions (2) and (1) we got:

$1066-588 =\mu +3\sigma -\mu +3\sigma$

$478= 6\sigma$

$\sigma =\frac{478}{6}=79.667$

3 0

3 years ago