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3 years ago

Can someone help me answer this?

Mathematics

1 answer:

AlekseyPX3 years ago

5 0

Answer:

statement number 3

Step-by-step explanation:

modes are both 1

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What the percentage of 3/5

fiasKO [112]

3 divide by 5 is 0.6 and you move the decimal two places to the right and your percentage is 60%

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3 years ago

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What is the value of a in the equation? 22= a/11

OLga [1]

Answer:

Step-by-step explanation:

If you divide 33 by 11 it’ll be 3 so we can cross B off, and A too because it’s not a double digit

So that leaves us with C & D.

If we divided 222 and 11 it would leave us with 10 which doesn’t match 11, so that leaves us with D

(Sorry if I didn’t’t explain right )

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3 years ago

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0.139 times what number could get me to 100? PLEASE ANSWER, FAST

Triss [41]

Answer:

719.4244604

Step-by-step explanation:

100/0.139=719.4244604

7 0

3 years ago

2r2 + 3s3 − r2 + 4t2 − r2 if r = −2, s = −3, and t = 5.

Elza [17]

First, simplify the expression $2r^2 + 3s^3 - r^2 + 4t^2 - r^2.$ Here you have three terms with $r^2,$ group them in the following way:

$2r^2 + 3s^3 - r^2 + 4t^2 - r^2=2r^2-r^2-r^2+3s^3+4t^2=(2r^2-r^2-r^2)+3s^3+4t^2=$

$=r^2(2-1-1)+3s^3+4t^2=3s^3+4t^2.$

Then substitute s=-3 and t=5:

$3s^3+4t^2=3\cdot (-3)^3+4\cdot 5^2=3\cdot (-27)+4\cdot 25=-81+100=19.$

Answer: 19.

3 0

3 years ago

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A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

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".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

5 0

4 years ago