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antiseptic1488 [7]

2 years ago

5

Helpppppppppp meeee please

1 answer:

Artemon [7]2 years ago

3 0

Answer:

<2 and <8

<3 and <5

Step-by-step explanation:

The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal , the resulting alternate interior angles are congruent .

This is my 100th question answered!

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Ede4ka [16]

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Physical education

Step-by-step explanation:

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3 years ago

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ANTONII [103]

Answer:

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Step-by-step explanation:

Hi!

An exponent is the same thing as just multiplying the expression by itself the number of times the exponent says. So we need to multiply 1/3 by itself three times.

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5 0

3 years ago

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May I please receive help on this question

skelet666 [1.2K]

Y+60=180 then y=120
5z-100=60 then 5z=160 then z=32

6 0

2 years ago

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Please help me with 1,2,3,4 and show me how you get it the answers

Dafna11 [192]

Answer:

Step-by-step explanation:

Use the formula of distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ ,

Distance = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

1. Distance between A(3, 4) and B(8, 12),

AB = $\sqrt{(8-3)^2+(12-4)^2}$

AB = $\sqrt{25+64}$

= $\sqrt{89}$

≈ 9.43

2. Distance between C(-2, 10) and D(3, -2),

CD = $\sqrt{(3+2)^2+(-2-10)^2}$

= $\sqrt{25+144}$

= $\sqrt{169}$

= 13.00

3. Distance between X(20, 4) and Y(17, 0),

XY = $\sqrt{(20-17)^2+(4-0)^2}$

= $\sqrt{3^2+4^2}$

= $\sqrt{25}$

= 5.00

4. Distance between E(2, -1) and M(6, 9),

EM = $\sqrt{(2-6)^2+(-1-9)^2}$

= $\sqrt{16+100}$

= $\sqrt{116}$

= 10.77

8 0

3 years ago

In a certain lake, trout average 12 in. in length with standard deviation 2.75 in. and the bass average 4 lb. in weight with sta

NISA [10]

Answer:

The bass fish was the better catch

Step-by-step explanation:

From the question we are told that

The population mean for trout is $\mu_1 = 12 \ in$

The standard deviation is $\sigma_1 = 2.75 \ in$

The population mean for base is $\mu _2 = 4 \ lb$

The standard deviation is $\sigma_2 = 0.8 \ lb$

The number of trout caught $x_1 = 18$

The number of bass caught $x_2 = 6$

Generally z-value(standardized value ) for the of number trout caught is mathematically represented as

$z_1 = \frac{x_1 - \mu_1}{\sigma_1 }$

substituting value

$z_1 = \frac{18 - 12}{2.75 }$

$z_1 = 2.18$

Generally z-value(standardized value ) for the of number bass caught is mathematically represented as

$z_2 = \frac{x_2 - \mu_2}{\sigma_2 }$

substituting value

$z_2 = \frac{6 - 4}{0.8 }$

$z_2 = 2.5$

From our calculation we see that $z_2 > z_1$

The fish that was the better catch is the bass fish

3 0

2 years ago