The output is ten less than the input

Can someone answer this it’s urgent

RideAnS [48]

Answer:

I don`t know about IQR but...

A:

Range = 50

Median = 105

Minimum = 80

Maximum = 130

B: 75%

C: 50%

D: 25%

E: 130%

Step-by-step explanation:

It izz wat it izzzz!!!

7 0

3 years ago

A number, x, rounded to 1 decimal place is 3.7<br>Write down the error interval for x.

alekssr [168]

Answer:

$3.65 \leqslant x \leqslant 3.75$

Step-by-step explanation:

The error in rounding a number is half of the unit of measure.

The number was rounded to the nearest 0.1 unit so the error is 1/2×0.1, which equals 0.05.

Now we add 3.7 and 0.05, which equals 3.75 and we also take 3.7 - 0.05, which equals 3.65

So, the error interval is:

$3.65 \leqslant x \leqslant 3.75$

4 0

3 years ago

Find the zero of: f(x)=5x+195

monitta

Answer:

(x + 39) or x = -39

Step-by-step explanation:

Set f(x) to 0

0 = 5x + 195

5x = -195

x = -39

4 0

2 years ago

Add the followinf expresion<br>b. x- 3y + 4z , y – 2x- 8z ,5x – 2y – 3z

Mariana [72]

Answer:

4x - 4y - 7z

Step-by-step explanation:

x - 3y + 4z + y - 2x - 8z + 5x - 2y - 3z =

(x - 2x + 5x) + (y - 3y - 2y) + (4z - 8z - 3z) =

4x + (- 4y) + (-7z) =

4x - 4y - 7z <===

3 0

3 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

5 0

3 years ago