All categories

3 years ago

What is the value of x in the figure? 60° (2x)

Mathematics

2 answers:

blagie [28]3 years ago

8 0

Answer:

The term X is a variable that takes the place of any number. ... Similarly 2 multiplied by X results to X plus X which means a number added to itself since multiplication is a repeated addition. So 2X = 2 * X = X + X = 2X.

Step-by-step explanation:

can i have brainliest

Zigmanuir [339]3 years ago

8 0

Answer:

x=30

Step-by-step explanation:

Since both of the acute angles have around or similar degrees, they will be the same. But even without that information we can see that each of the angles look the same. We already know that there are 360° in a circle, making 180° in a semi-circle. We can add 60+60 which will give us 120°. 360-120=240, which dividing that by 2 will give us 120, and if we were figuring out the other 2 would be 120° each.

You might be interested in

Which number is equivalent to 34/32

Anestetic [448]

34/32 can also be written as 1.0625 or 1 & 1/16.

8 0

3 years ago

ILL GIVE U BRAINLYIST

Amiraneli [1.4K]

B the person above me is right

7 0

2 years ago

PLEASE HEEELPPP 15 POINTS PLESSEE

sergeinik [125]

Answer:

200.96 yd^2

Step-by-step explanation:

Area of a circle = πr^2 where r = radius

radius = diameter/2

radius = 16/2

radius = 8

Area = π(8)^2

Area = π64

Area = 200.96 yd^2

6 0

3 years ago

Read 2 more answers

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago

Alice is playing her bill at a restaurant. The tax on the cost of her meal is 5%.She decides to leave a tip of 20% of her cost o

Oxana [17]

Answer:

5% * 20% = 0.01

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers