What is the estimated probability that at least two of the puppies will be

An Internet store charges a base fee plus $1.50 per pound to ship items. The table shows samples of their shipping rates.

nalin [4]

Answer: For Last Question 4.50

Step-by-step explanation:It depends on the price per pound. Let's say the price per pound is 50 cents (or 0.50). (This is an example, the answer I'm giving will most likely NOT be correct, due to the nature of your question - sorry!) To calculate the price of Derek's package, you would multiply 7 (the weight of the package) by 0.50 (the price per pound), which totals to 3.50 Next, to calculate the price for Keisha's package, you'd do the same thing, but instead of multiplying 7 by the price per pound, you would multiply 10 by the price per pound, or 0.50. This multiplies out to 5 dollars. Next you'd subtract Derek's price from Keisha's, which is 5 - 3.50 or 1.50. If it is asking for a percentage, it would be about 150%.

7 0

3 years ago

Read 2 more answers

What is x-35>15? plz help

alexira [117]

Answer: x > 50

Step-by-step explanation:

5 0

2 years ago

12 + 3 to the power of 2 + 50<br><br> (using PEMDAS)

Musya8 [376]

First you do 3 to the 2nd power which is 9 then you just add 12+9+50= 71

8 0

3 years ago

Read 2 more answers

The NCAA is interested in estimating the difference in mean number of daily training hours for men and women athletes on college

hammer [34]

Answer:

$s^2_p = \frac{9*0.3^2 +9*0.4^2}{10+10-2}=0.125$

$s_p =\sqrt{0.125}=0.354$

$(2.7 -2.4) - 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=-0.032$

$(2.7 -2.4)+ 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=0.632$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Data given

$\bar X_M = 2.7$ represent the sample mean for men

$\bar X_F = 2.4$ represent the sample mean for women

$s_M = 0.3$ represent the sample deviation for men

$s_F = 0.4$ represent the sample deviation for women

$n_M = 10$ sample size of male

$n_F =10$ sample size of women

The confidence interval is given by:

$(\bar X_M -\bar X_F) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_M}+\frac{1}{n_F}}$ (1)

The polled variance can be calculated with this formula:

$s^2_p = \frac{9*0.3^2 +9*0.4^2}{10+10-2}=0.125$

$s_p =\sqrt{0.125}=0.354$

For a confidence of 95% the value for the significance is $\alpha=1-0.95=0.05$ and $\alpha/2 = 0.025$ , the degrees of freedom are given by:

$df = n_M + n_F -2= 10+10-2=18$

And the critical value can be calculated with the following formula in excel: "=T.INV(1-0.025,18)" and we got $t_{\alpha/2}=2.1$

Now we can replace into the confidence interval:

$(2.7 -2.4) - 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=-0.032$

$(2.7 -2.4)+ 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=0.632$

6 0

3 years ago

Write in standard form 2+4i over 1+i

Levart [38]

So hmm a conjugate, is pretty much just, the same binomial, but with a different sign in the middle, so, a + b, has a conjugate of a - b
or -a + b, has a conjugate of - a - b, or c - d, has a conjugate of c +d, and so on

anyway, the idea being, to "rationalize" the expression, namely, getting rid of the pesky radical in the denominator

so, we'll multiply the expression by 1, since anything times 1 is just itself

however, bear in mind, that 1, can be a/a, or b/b, or cheese/cheese, or anything/anything

so, we'll multiply the top and bottom of the fraction, by the conjugate of the denominator

anyhow, that said $\bf \cfrac{2+4i}{1+i}\cdot \cfrac{1-i}{1-i}\implies \cfrac{(2+4i)(1-i)}{(1+i)(1-i)}\implies \cfrac{2-2i+4i-4i^2}{1-i^2}\\\\
-----------------------------\\\\
\textit{recall that }i^2=-1\\\\
-----------------------------\\\\
 \cfrac{2-2i+4i-4(-1)}{1-(-1)}\implies \cfrac{2+2i+4}{2}\implies \cfrac{6+2i}{2}\implies \cfrac{6}{2}+\cfrac{2i}{2}
\\\\\\
\boxed{3+i}$

7 0

2 years ago