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3 years ago

12

You’re given two side lengths of 6 centimeters and 9 centimeters. Which measurement can you use for the lengths of the third sid

e to construct a valid triangle

1 answer:

Vadim26 [7]3 years ago

3 0

If we have a triangle with sides a,b,c then

b-a < c < b+a

describes the restriction of the length of side c, given sides a and b. This is due to the triangle inequality theorem.

a = 6

b = 9

b-a < c < b+a

9-6 < c < 9+6

3 < c < 15

<h3>Answer: The third side is between 3 cm and 15 cm. The third side cannot equal 3 cm, and it also cannot equal 15 cm. </h3>

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lesya692 [45]

The student would earn %86. To calculate percentage you take the amount of questions correct, divide that by the amount of questions given, multiply by 100 and round to the nearest percent.

3 0

3 years ago

In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n = 50

sample mean $\overline x$ = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

$X \sim N ( \mu \sigma)$

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})$

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})$

$P(X>140) = 1 - P( Z\leq0.42)$

From z tables,

$P(X>140) = 1 - 0.6628$

$P(X>140) = 0.3372$

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

$Y \sim Binomial (np)$

$Y \sim Binomial (50,0.3372)$

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

$P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}$

P(Y ≥ 15) = 0.763

7 0

3 years ago

A sample contains 400 Plutonium-244 atoms. After 1 half-life(s) how many plutonium atoms are left on average?

SpyIntel [72]

Answer:

200

Step-by-step explanation:

Divide 400 by 1/2 or 2.

8 0

3 years ago

A trough is 12 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1

V125BC [204]

Answer:

$\frac{dh}{dt}=0.5ft$

Step-by-step explanation:

From the question we are told that:

Length $l=12$

Top length $l_t=3ft$

Height $h=1ft$

Rate $R=14 ft3/min$

Water rise $w=4$

Generally the equation for Velocity is mathematically given by

$V=frac{1}{2}wh'(l)\\\\V=frac{1}{2}wh'(12)$

$V=18h'^2$

Therefore

$R=18(2h)(\frac{dh}{dt})$

Where

$h=\frac{3}{4}$

Therefore

$\frac{dh}{dt}=\frac{R}{18(2h)}$

$\frac{dh}{dt}=\frac{14}{18(2.3/4)}$

$\frac{dh}{dt}=0.5ft$

8 0

3 years ago

Using a proportion, what can you infer about the number of households in her town that have more than three children?

eduard

Answer:

About 296 households have more than three children.

Step-by-step explanation:

The complete question is:

Carmen used a random number generator to simulate a survey of how many children live in the households in her town. There are 1,346 unique addresses in her town with numbers ranging from zero to five children. The results of 50 randomly generated households are shown below. Children in 50 Households Number of Children Number of Households 0 5 1 11 2 13 3 10 4 9 5 2 Using a proportion, what can you infer about the number of households in her town that have more than three children? About 242 households have more than three children. About 269 households have more than three children. About 296 households have more than three children. About 565 households have more than three children.

Solution:

The data provided for the number of children and number of households is:

Number of Children (<em>X</em>) Number of Households (<em>f</em> (X))

0 5

1 11

2 13

3 10

4 9

5 2

TOTAL 50

Compute the probability of households having more than three children as follows:

P (More than 3 children) = P (4 children) + P (5 children)

$=\frac{9}{50}+\frac{2}{50}$

$=\frac{11}{50}$

$=0.22$

Let the random variable <em>Y</em> be defined as the number of households having more than three children.

The probability of the random variable <em>Y </em>is, <em>p</em> = 0.22.

The entire population, i.e. total number of addresses in Carmen's town with numbers ranging from zero to five children, consists of <em>N</em> = 1,346 households.

Compute the expected number of households having more than three children as follows:

$E(Y) =N\times p$

$=1346\times 0.22\\=296.12\\\approx 296$

Thus, about 296 households have more than three children.

4 0

3 years ago