What is the result when the number 25 is increased by 20

(2x30)-6 help me out

taurus [48]

Answer:

54

Step-by-step explanation:

2 x 30 = 60

60 -6 = 54

6 0

2 years ago

Read 2 more answers

A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

5 0

2 years ago

Ratios that show the same relationship are

olchik [2.2K]

Equivalent because they are the same

8 0

3 years ago

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A pool company is creating a blueprint for a family pool and a similar dog pool for a new client. Which statement explains how t

Charra [1.4K]

The statement that explains how the company can determine whether pool LMNO is similar to pool PQRS is;

B. Translate PQRS so that point Q of PQRS lies on point M of LMNO, then dilate PQRS by the ratio segment PQ over segment LM.

<h3>How to carry out Transformations?</h3>

Given that quadrilaterals ABCD and EFGH are similar:

The corresponding points on the quadrilaterals are:

P → L

Q → M

R → N

S → O

So, the first step is any of the following:

Translate point P to L

Translate point Q to M

Translate point R to N

Translate point S to O

Notice that the side lengths of PQRS are bigger than that of LMNO

This means that the Quadrilateral PQRS has to be dilated (compressed) by a ratio of side lengths of LMNO divided by side lengths of PQRS

For example, the point M is translated to point Q. The figure will then be dilated by a ratio of LM divided by PQ.

Read more about Transformations at; brainly.com/question/4289712

#SPJ1

3 0

1 year ago

Solving Quadratic Equations using the Square Root Property

Afina-wow [57]

Answer:

$x = -2$ or $x = -5$

Step-by-step explanation:

You need to complete the square before you can take the square root of both sides.

$x^2 + 7x + 10 = 0$

Subtract 10 from both sides.

$x^2 + 7x = -10$

To complete the square, you need to add the square of half of the x-term coefficient to both sides.

The x-term coefficient is 7. Half of that is 7/2. Square it to get 49/4. Now we add 49/4 to both sides of the equation.

$x^2 + 7x + \dfrac{49}{4} = -10 + \dfrac{49}{4}$

$(x + \dfrac{7}{2})^2 = -\dfrac{40}{4} + \dfrac{49}{4}$

$(x + \dfrac{7}{2})^2 = \dfrac{9}{4}$

Now we use the square root property, if

$x^2 = k$ , then

$x = \pm \sqrt{k}$

$x + \dfrac{7}{2} = \pm \sqrt{\dfrac{9}{4}}$

$x + \dfrac{7}{2} = \pm \dfrac{3}{2}$

$x + \dfrac{7}{2} = \dfrac{3}{2}$ or $x + \dfrac{7}{2} = -\dfrac{3}{2}$

$x = -\dfrac{4}{2}$ or $x = -\dfrac{10}{2}$

$x = -2$ or $x = -5$

5 0

3 years ago