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Annette [7]
3 years ago
9

mr. cuddy draws a triangle with a perimeter of 36cm. Principal Aranda says that the longest side measures 18 cm, How do you know

that the principal Aranda is incorrect? Explain
Mathematics
1 answer:
hichkok12 [17]3 years ago
3 0

Answer:

See the procedure

Step-by-step explanation:

we know that

<u>The Triangle Inequality Theorem</u>, states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

Let

a,b,c the lengths side of triangle

c is the greater side

The perimeter is equal to

P=a+b+c

P=36 cm

If c=18 cm

then

a+b=18

Applying the Triangle Inequality Theorem

a+b > c

18 > 18  ----> is not true

therefore

Principal Aranda is incorrect

The larger side cannot measure 18 cm

The largest side must be less than 18 cm

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A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observationa
diamong [38]

Answer:

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}= \mu_D

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C,D

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have p=4 groups and on each group from j=1,\dots,p we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2  

And we have this property  

SST=SS_{between}+SS_{within}  

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.

8 0
3 years ago
Find the Area of the figure below, composed of a parallelogram and two semicircles. Round to the nearest tenths place.
Mila [183]

Step-by-step explanation:

the total area = the area of full circle + the area of parallelogram = (π ×(8/2)²) + (28×7)

= (3,14 ×16)+(196)

= 50,24 +196

= 246,24

= 246,2

7 0
2 years ago
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
Suppose f(x) = 3x2 – 2x: Find f(-4). А B -40 ОО C 56<br>plz helpp​
amid [387]

Answer:

f(-4) = 56

Step-by-step explanation:

f(x) = 3x² - 2x

To find f(-4), substitute -4 for all the values of x in f(x).

f(-4) = 3(-4)² - 2(-4)

Square -4.

f(-4) = 3(16) - 2(-4)

Multiply 3 and 16.

f(-4) = 48 - 2(-4)

Multiply -2 and -4.

48 + 8

Add.

56

You had the right idea! :)

7 0
3 years ago
Square ABCD is shown with four congruent images such that ABCD FGHI JKLM NOPQ RSTU.
Anettt [7]

Answer:

  • <em><u>RSTU</u></em>

Explanation:

The question states the <em>congruent</em> images:

  • ABCD ≅ FGHI ≅ JKLM ≅ NOPQ ≅ RSTU

The order of the letters matters.

For instance, ABCD ≅ FGHI means that the vertex A is transformed into vertex F, the vertex B is transformed into the vertex G, the vertex C is transformed into vertex H, and the vertex D is transformed into vertex I.

From the images, the square ABCD is just shifted 9 units down to form the square FGHI. No refelection is needed.

As per the square RSTU only a reflection of square ABCE accross the y-axis makes the vertex A into vertex R, the vertex B into the vertex S, the vertex C into the vertex T, and the vertex D into the vertex U.

3 0
3 years ago
Read 2 more answers
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