Will mark brainliest

Write the equation of the line that passes through the points (2 , -1) and has a slope of - 3

svetlana [45]

Answer:

y=-3x+5

Step-by-step explanation:

y=mx+b

m=slope or gradient (=-3)

y=-3x+b

Add your points

(2, -1)

-1=(-3 x 2)+b

-1= -6+b

-1+6=b

b=5

y=-3x+5

5 0

3 years ago

Read 2 more answers

You volunteer to help drive children at a charity event to the zoo, but you can fit only 7 of the 18 children present in your va

yulyashka [42]

The answer is 19,448 different groups, using the formula from permutation and combinations C (n, r) = (n!) / (r! (n-r)!) which calculates the number of times where r objects can be chosen from n object.

So,

C (17,7) = (17!) / (7! *(17-7)!)

C (17,7) = (17!) / (7! *10!)

C (17,7) = (17*16*15*14*13*12*11*10!) / (7! *10!) then eliminate 10!

C (17,7) = (17*16*15*14*13*12*11) / (7*6*5*4*3*2*1) = (98017920)(5040)

C (17,7) = 19,448 different groups

3 0

4 years ago

An umbrella is made by stitching 10 triangular pieces of clothes of 2 different colors, each piece measuring 9 cm, 40 cm and 41

malfutka [58]

Answer:

See below in bold.

Step-by-step explanation:

The triangular piece of cloth is a right triangle because 41^2 = 40^2 + 9^2 (By the Pythagoras theorem).

So the area of one piece = 1/2 * 9 * 40 cm^2.

The total amount of cloth required for one umbrella = 10 * 1/2 * 9 * 40

= 1800 cm^2.

The total cost = 1800 * 0.25 = Rs 450.

3 0

3 years ago

Make a friend right now 10 points for whoever douse

Afina-wow [57]

Hand me dem points rn

5 0

4 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