What is the value of x in the equation 0.02x+0.7=0.8−0.03x ?

What is the smaller super script above the base that indicates how many times the

Alenkasestr [34]

I believe it’s the power

8 0

3 years ago

Read 2 more answers

Please I need help ASAP

jasenka [17]

Answer:

A

Step-by-step explanation:

3 0

3 years ago

Evaluate the expression -2

DaniilM [7]

Answer: 2x for x=3: 2(3) = 6.

Step-by-step explanation:

6 0

3 years ago

You are going to estimate how long a person holds down a button on their calculator by randomly selecting people and asking them

Ray Of Light [21]

Answer:

You need to have 538 people in your study.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.94}{2} = 0.03$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.03 = 0.97$ , so Z = 1.88.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

You practiced on the people at work and found a standard deviation of about 0.37 seconds.

This means that $\sigma = 0.37$

You want to get a 94% confidence interval that is only 0.06 in width.

This means that $M = \frac{0.06}{2} = 0.03$

How many people do you need to have in your study?

This is n for which M = 0.03. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.03 = 1.88\frac{0.37}{\sqrt{n}}$

$0.03\sqrt{n} = 1.88*0.37$

$\sqrt{n} = \frac{1.88*0.37}{0.03}$

$(\sqrt{n})^2 = (\frac{1.88*0.37}{0.03})^2$

$n = 537.6$

Rounding up:

You need to have 538 people in your study.

7 0

3 years ago

To determine whether or not they have a certain desease, 304 people are to have their blood tested. However, rather than testing

Artyom0805 [142]

Answer:

The expected number of tests necessary for the entire group of 304 people is 180.0325

Step-by-step explanation:

Probability of person being positive = p = 0.04

Probability of person being negative = q = 1-0.04=0.96

Number of people in each group = 19

We will use binomial over here

Probability of no test positive :

$P(x=0) = ^{19}C_0 (0.04)^0 (0.96)^{19} =0.460419201958$

P( at least one tests positive) = 1-P( no one tests positive)= 1-0.460419201958= 0.539580798042

Expected number of tests for each group = 1(0.460419201958)+20(0.539580798042) = 11.2520351628

Number of groups = $\frac{\text{Total population}}{\text{no. of people in each group}}=\frac{304}{19}=16$

Expected number of tests necessary for the entire group of 304 people:

= $\text{Expected number of tests for each group} \times \text{No. of groups}$

= $11.2520351628 \times 16$

=180.0325

Hence The expected number of tests necessary for the entire group of 304 people is 180.0325

7 0

3 years ago