Select three ratios that are equivalent to 4:18

Please help me really quick pleasee answer choices: a)2 b)11 c)20 d)29

mihalych1998 [28]

Answer:

C. $x=20^{\circ}$

Step-by-step explanation:

Angle 6 and angle with measure 45° are vertical angles (opposite angles whan two lines intersect). By Vertical Angles Theorem, vertical angles are congruent, so

$m\angle 6=45^{\circ}$

Angle 6 and angle with measure $5x+35^{\circ}$ are the same side interior angles when two parallel lines are cut by transversal. The same side interior angles add up to 180°, then

$m\angle 6+5x+35^{\circ}=180^{\circ}\\ \\45^{\circ}+5x+35^{\circ}=180^{\circ}\\ \\5x=180^{\circ}-45^{\circ}-35^{\circ}\\ \\5x=100^{\circ}\\ \\x=20^{\circ}$

7 0

3 years ago

Find the value of a and b if root3-1/root3+1+ root3+1/root3-1=a+root3 b

Digiron [165]

Answer:

a=4 and b = 0

Step-by-step explanation:

Given : $\dfrac{\sqrt 3 -1}{\sqrt 3+1}+\dfrac{\sqrt 3+1}{\sqrt 3-1}=a+\sqrt 3 b$

To find, The value of a and b

Solution,

Solving LHS of the given equation,

$\dfrac{(\sqrt 3-1)^2+(\sqrt3+1)^2}{(\sqrt 3+1)(\sqrt 3-1)}=a+\sqrt 3 b$

Since,

$(a-b)^2=a^2+b^2-2ab\\\\(a+b)^2=a^2+b^2+2ab\\\\(a-b)(a+b)=a^2+b^2$

So,

$\dfrac{3+1-2\sqrt 3+3+1+2\sqrt 3}{3-1}=a+\sqrt 3 b\\\\4=a+\sqrt 3 b$

or

$4+0=a+\sqrt 3 b$

On comparing we get :

a = 4 and b = 0

4 0

3 years ago

Solve ax + bx = c for x a,b,c represent nonzero constants

Zepler [3.9K]

Answer:

x=(c/a+b)

Step-by-step explanation:

..........

4 0

3 years ago

Read 2 more answers

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0

3 years ago

A square has side length of 5 cm. The square was changed using a scale factor of 0.6.

Aliun [14]

So the side length became 0.6*5 = 3 cm.

8 0

3 years ago