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Studentka2010 [4]

2 years ago

11

A rectangle made of square tiles measures 10 tiles long and 8 tiles wide. what is the width of a similar rectangle whose length

is 15 tiles?

2 answers:

ahrayia [7]2 years ago

5 0

Answer:

13

Step-by-step explanation:

postnew [5]2 years ago

4 0

Answer:

12 or 13

hope this helps

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If A = (-2, -4) and B = (-8, 4), what is the length of AB?

svetoff [14.1K]

Answer: OPTION D

Step-by-step explanation:

To solve this exercise you must use the formula for calculate the distance between two points, which is shown below:

$d_{(A,B)}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Now, you must substitute the points given in the problem into the formula:

A(-2,-4)

B(-8,4)

$d_{(A,B)}=\sqrt{(-8-(-2))^2+(4-(-4))^2}$

Then, the result is:

$d_{(A,B)}=10units$

5 0

2 years ago

Read 2 more answers

If 2x3 – 4x2 + kx + 10 is divided by (x + 2), the remainder is 4. Find the value of k using remainder theorem. Please help :)

Vinvika [58]

The polynomial remainder theorem states that the remainder of the division of a polynomial $P(x)$ by $x-a$ is equal to $P(a)$ .

Therefore

$P(-2)=4\\2\cdot(-2)^3 - 4\cdot(-2)^2 + k\cdot(-2) + 10=4\\-16-16-2k=-6\\-2k=26\\k=-13$

8 0

2 years ago

Twenty five is the quotient of a number y and 3.5

Vilka [71]

Y/3.5= 25

(Hoped this help^)

8 0

2 years ago

Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of

Alexxx [7]

Answer:

a) 0.018

b) 0

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected men, these men have a mean hip breadth greater than 16.5 in)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)$

$= 1 - P(z \leq 23.29)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 1 = 0$

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0

2 years ago

Simplify: (g^6)(g^11)

11Alexandr11 [23.1K]

Multiplying exponents with like bases is the same as adding up the exponents.
<span>(g^6)(g^11)=g^17

Hope this helps!</span>

4 0

2 years ago