-m + 10 = -8 – 6m + 3m

The product of 1 1/2 and 2 is less than, equal to, or greater than 2

Rus_ich [418]

Answer:

The answer is 1 1/2 < 2

Step-by-step explanation:

If we convert 1/2 into a decimal, we would get .5. Add that to 1 and we get 1.5. Therfore, 2 > 1.5

3 0

3 years ago

Find the 95% confidence interval for estimating the population mean μ

AVprozaik [17]

We first need to determine whether we are dealing with means or proportions in this problem. Since we are given the sample and population mean, we know that we are dealing with means.

Since we have one sample mean, this means we are creating a confidence interval for one sample (1 Samp T Int).

Normally we would check for conditions, but since this is not formulated as a "real-world scenario" type problem, it is hard to check for randomness and independence. Therefore, I will be excluding conditions from this answer.

<h3>Confidence Interval Formula</h3>

The formula for constructing a <u>confidence interval for means</u> is as follows:

$\displaystyle \overline{x} \pm t^*\big{(}\frac{\sigma}{\sqrt{n} } \big{)}$

We are given these variables:

$\overline{x}=50$
$n=60$
$\sigma=10$

Plug these values into the formula for the confidence interval:

$\displaystyle 50\pm t^* \big{(}\frac{10}{\sqrt{60} } \big{)}$

<h3>Finding the Critical Value (t*)</h3>

In order to find t*, we can use this formula:

$\displaystyle \frac{1-C}{2}=A$

Calculating the z-score associated with "A" will give us t*.

So, let's plug in the confidence interval 95% (.95) into the formula:

$\displaystyle \frac{1-.95}{2}=.025$

Use your calculator or a t-table to find the z-score associated with this area under the curve.. you should get:

$t^*=1.96$

<h3>Constructing Confidence Interval</h3>

Now, let's finish the confidence interval we created:

$\displaystyle 50\pm 1.96 \big{(}\frac{10}{\sqrt{60} } \big{)}$

We can calculate the confidence interval, using this formula, to be:

$\boxed{(47.4697, \ 52.5303)}$

<h3>Interpreting the Confidence Interval</h3>

We are 95% confident that the true population mean μ lies between <u>47.4697 and 52.5303</u>.

8 0

2 years ago

Elle decreased her monthly budget from $250 to $200. Calculate the percent decrease

kirill [66]

Formula for finding percent change (increase or decrease) is amount of the change divided by the original amount.

Here, the amount of the change is 50 and the original amount is 250.

50/250= 0.2

Multiply by 100 to get percent.

20% decrease (because budget went down)

6 0

3 years ago

Travis bought two plots of land at the beginning of 2000. When purchased, the smaller plot of land was valued at $20,000.00 and

stepan [7]

HAT MATH ARE U IN
i got part of it but it lost me the
<span>differ part of land is 49,500 but it lost meh hope it helps gl</span>

3 0

3 years ago

Read 2 more answers

Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ

-Dominant- [34]

Answer:

$\tan(\beta)$

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something. If you have options that you're building toward, aim toward one of them.

${\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}$ and ${\sec(\theta)}={\dfrac{1}{\cos(\theta)}$

Recall the following reciprocal identity:

$\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}$

So, the original expression can be written in terms of only sines and cosines:

$\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)$

$\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }$

$\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}$

$\sin(\beta) *\dfrac{1 }{\cos(\beta) }$

$\dfrac{\sin(\beta)}{\cos(\beta) }$

Working toward one of the answers provided, this is the tangent function.

The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero. However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to $\tan(\beta)$ .

8 0

2 years ago