275 per 60 sec divided by 7

Point M begins at (5,-2). Write the coordinates of Ry= 1 (M). Is this correctly illustrated

yawa3891 [41]

Answer:

No

Step-by-step explanation:

The reflection (R) around the line y = 1 is not the point illustrated.

Notice that the original point M (5,-2) is THREE (3) units below the line y=1, therefore when doing the reflection, it should end up THREE units above the y=1 line.

That is, it should end up at the point (5,4 ) and not at (5,3) as is shown in the picture.

7 0

3 years ago

Find the value of x in each case:<br><br>(Will give brainliest)

Novay_Z [31]

Answer:

x= 22.5°

Step-by-step explanation:

∠DEA= ∠BAE (alt. ∠s, DE// AB)

Substitute ∠DEA= 2x:

∠BAE= 2x

∠AEB +∠BEF= 180° (adj. ∠s on a str. line)

Substitute ∠BEF= 4x:

∠AEB +4x= 180°

∠AEB= 180° -4x

∠ABE +∠CBE= 180° (adj. ∠s on a str. line)

Substitute ∠CBE= 6x:

∠ABE +6x= 180°

∠ABE= 180° -6x

∠BAE +∠AEB +∠ABE= 180° (∠ sum of triangle)

2x +180° -4x +180° -6x= 180°

-8x +360°= 180°

8x= 360° -180°

8x= 180°

x= 180° ÷8

x= 22.5°

6 0

3 years ago

Learning Task 1: Translate each of the following problem. Write your an swer in your notebook. 1. The quotient of $2 and 2 8 2.

finlep [7]

Answer:

1. 78/5

2. 12

3. 25/8

4. 48/35

5. 21/2

6 0

3 years ago

Which of the functions graphed below is continuous?

vaieri [72.5K]

Answer:

The answer is the second one

5 0

3 years ago

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experime

Citrus2011 [14]

Answer:

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

Step-by-step explanation:

We have to perform a hypothesis test on the difference between means.

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2$

μ1: mean heat output for subjects with the syndrome.

μ2: mean heat output for non-sufferers.

We will use a significance level of 0.05.

The difference between sample means is:

$M_d=\bar x_1-\bar x_2=0.63-2.09=-1.46$

The standard error is

$s_{M_d}=\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}=\sqrt{0.3^2/9+0.5^2/9}=\sqrt{ 0.038 } \\\\ s_{M_d}=0.194$

The t-statistic is

$t=\dfrac{M_d}{s_{M_d}}=\dfrac{-1.46}{0.194}=-7.52$

The degrees of freedom are

$df=n_1+n_2-2=9+9-2=16$

The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.

The t-statistic is below the critical value, so it lies in the rejection region.

The null hypothesis is rejected.

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

5 0

3 years ago