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3 years ago

5

Use the balanced scale to find the conversion factor that can be used to convert the number of blocks to the weight of the block

s in pounds.

1 answer:

Alexandra [31]3 years ago

6 0

Find Total weight at left side

$\\ \sf\longmapsto 6(1lb)=6lbs$

Now

Per block weight at right side

$\\ \sf\longmapsto \dfrac{6}{4}=1.5lbs$

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I need help with DE Precalculus, mainly trig identities. Would anyone be willing to zoom or text me to help tutor? Thanks!

liq [111]

Answer:

I would help but I'm really tired

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3 years ago

What would the variables for a and b be? 3/x + 5/x^2 = (ax+b)/x^2

allsm [11]

Answer:

a=3,b=5

Step-by-step explanation:

$\frac{3}{x} + \frac{5}{ {x}^{2} } \\ = \frac{3x}{ {x}^{2} } + \frac{5}{ {x}^{2} } \\ = \frac{3x + 5}{ {x}^{2} }$

so as you got it:

$\frac{(ax + b)}{ {x}^{2} } = \frac{3x + 5}{ {x}^{2} } \\ a = 3 \\ b = 5$

4 0

3 years ago

You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample

Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=160)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{1.96(160)}{78.4})^2 =16$

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And now we can replace the new value of Me and see what we got, like this:

$n=(\frac{1.96*160}{60})^2 =27.32$

And if we round up the answer we see that the value of n to ensure the margin of error required $Me=\pm 60$ $ is n=28.

5 0

3 years ago

Suppose we have an unfair coin that its head is twice as likely to occur as its tail. a)If the coin is flipped 3 times, what is

Alenkinab [10]

Answer: a) 0.2222, b) 0.3292, c) 0.1111

Step-by-step explanation:

Since we have given that

Let the probability of getting head be p.

Since, its head is twice as likely to occur as its tail.

$p+\dfrac{p}{2}=1\\\\\dfrac{3p}{2}=1\\\\p=\dfrac{2}{3}$

a)If the coin is flipped 3 times, what is the probability of getting exactly 1 head?

So, here, n = 3

$p=\dfrac{2}{3}$

$q=\dfrac{1}{3}$

Now,

$P(X=1)=^3C_1(\dfrac{2}{3})^1(\dfrac{1}{3})^2=0.2222$

b)If the coin is flipped 5 times, what is the probability of getting exactly 2 tails?

2 tails means 3 heads.

So, it becomes,

$P(X=3)=^5C_3(\dfrac{2}{3})^3(\dfrac{1}{3})^2=0.3292$

c)If the coin is flipped 4 times, what is the probability of getting at least 3 tails?

$P(X\leq 1)=\sum _{x=0}^1^4C_x(\dfrac{2}{3}^x(\dfrac{1}{3})^{4-x}=0.1111$

Hence, a) 0.2222, b) 0.3292, c) 0.1111

8 0

3 years ago

To ________ a formula or equation for a specified variable means to rearrange it so that the variable appears alone on one side

777dan777 [17]

<span>I think they're looking for 'isolate' maybe..or it could be 'solve' </span>

4 0

3 years ago

Read 2 more answers