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2 years ago

What is 1345/99 rounded to the nearest integer

Mathematics

2 answers:

ludmilkaskok [199]2 years ago

7 0

1345/99 = 15.5858
15.5858 round up because 5+ is round up while 4- is round down.
Your answer is 16.

vesna_86 [32]2 years ago

4 0

Solutions

An integer is a whole number that can be positive, negative, or zero.

Integers = { ..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ... }

Numbers that are not integers = { 1.43, 1 3/4, 3.14, .09, and 5,643.1.... }

In mathematical equations, unknown or unspecified integers are represented by lowercase, italicized letters from the "late middle" of the alphabet.

1345/99 rounded to the nearest integer

1345/99 = 15.5858

Rounded answer = 16.

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Jade has $451.89 in her checking account. After 3 checks, each for the same amount, are deducted from her account, she has $358.

steposvetlana [31]

Answer:

3rd, 4th, and 5th options.

Step-by-step explanation:

The way this was formatted was a little weird, but the 3rd, 4th, and 5th options are correct I think.

If there was 358.92 left, it has to be divided 3 ways (after subtracting 451.89 - 358.92). So that means that 3c or c+c+c could be correct, as well as using dis. property to switch around the last one I guess.

if this helped, brainliest is appreciated.

4 0

2 years ago

Read 2 more answers

Show that the Fibonacci numbers satisfy the recurrence relation fn = 5fn−4 + 3fn−5 for n = 5, 6, 7, . . . , together with the in

Sonja [21]

Answer with step-by-step explanation:

We are given that the recurrence relation

$f_n=5f_{n-4}+3f_{n-5}$

for n=5,6,7,..

Initial condition

$f_0=0,f_1=1,f_2=1,f_3=2,f_4=3$

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

$f_n=f_{n-1}+f_{n-2}$ , $f_0=0,f_1=1$

Apply this

$f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}$

$f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}$

$f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}$

Substitute n=2

$f_2=f_1+f_0=1+0=1$

$f_3=f_2+f_1=1+1=2$

$f_4=f_3+f_2=2+1=3$

Hence, the Fibonacci numbers satisfied the given recurrence relation .

Now, we have to show that $f_{5n}$ is divisible by 5 for n=1,2,3,..

Now replace n by 5n

$f_{5n}=5f_{5n-4}+3f_{5n-5}$

Apply induction

Substitute n=1

$f_5=5f_1+3f_0=5+0=5$

It is true for n=1

Suppose it is true for n=k

$f_{5k}=5f_{5k-4}+3f_{5k-5}$ is divisible 5

Let $f_{5k}=5q$

Now, we shall prove that for n=k+1 is true

$f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)$

$f_{5k+5}=5(f_{5k+1}+3q)$

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the $f_{5n}$ is divisible by 5 for n=1,2,3,..

8 0

1 year ago

Triangle XYZ with vertices X(0, 0), Y(0, –2), and Z(–2, –2) is rotated to create the image triangle X'(0, 0), Y'(2, 0), and Z'(2

Marina86 [1]

Answer: The correct options are

(A) Rotation 90° anticlockwise.

(D) (x, y) → (–y, x).

Step-by-step explanation: Given that ΔXYZ is rotated to create the image triangle ΔX'Y'Z'.

Triangle XYZ and its image triangle X'Y'Z' are shown in the attached figure.

The co-ordinates of the vertices of ΔXYZ are X(0, 0), Y(0, -2) and Z(-2, -2).

And the co-ordinates of the vertices ΔX'Y'Z' are X'(0, 0), Y'(2, 0) and Z'(2, -2).

Option (A) Rotation 90°:

We see from the figure that if we rotate ΔXYZ is rotated 90° anticlockwise, then it will coincide with ΔX'Y'Z'.

So, rotation of 90° anticlockwise is a correct option.

Option (B) Rotation 180°:

If we rotate ΔXYZ is rotated clockwise or anticlockwise 180°, then it will NOT coincide with ΔX'Y'Z'.

So, rotation of 180° is NOT a correct option.

Option (C) Rotation 270°:

If we rotate ΔXYZ is rotated clockwise 270°, then also it will not coincide with ΔX'Y'Z'.

So, rotation of 270° clockwise is also a correct option.

Option (D) (x, y) → (–y, x):

We see that the co-ordinates of both the triangle follow the transformation

X(0, 0) ⇒ X'(0, 0)

Y(0, -2) ⇒ Y'(2, 0)

Z(-2, -2) ⇒ Z'(2, -2).

So, the transformation is (x, y) ⇒ (-y, x).

Therefore, the transformation (x, y) → (–y, x) is a correct option.

Option (E) (x, y) → (y, -x):

We see that the co-ordinates of both the triangle does NOT follow this transformation

For example, suppose this transformation is correct. Then, we have

Y(0, -2) ⇒ (-2, 0), which are not the co-ordinates of Y'.

Therefore, the transformation (x, y) → (–y, x) is NOT a correct option.

Thus, the correct options are:

(A) Rotation 90° anticlockwise.

(D) (x, y) → (–y, x).

9 0

1 year ago

Read 2 more answers

A toddler is allowed to dress himself on Mondays, Wednesdays, and Fridays. For each of his shirt, pants, and shoes, he is equall

avanturin [10]

Answer:

0.0286 = 2.86% probability that today is Monday.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Dressed correctly

Event B: Monday

Probability of being dressed correctly:

100% = 1 out of 4/7(mom dresses).

(0.5)^3 = 0.125 out of 3/7(toddler dresses himself). So

$P(A) = 0.125\frac{3}{7} + \frac{4}{7} = \frac{0.125*3 + 4}{7} = \frac{4.375}{7} = 0.625$

Probability of being dressed correctly and being Monday:

The toddler dresses himself on Monday, so (0.5)^3 = 0.125 probability of him being dressed correctly, 1/7 probability of being Monday, so:

$P(A \cap B) = 0.125\frac{1}{7} = 0.0179$

What is the probability that today is Monday?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0179}{0.625} = 0.0286$

0.0286 = 2.86% probability that today is Monday.

4 0

1 year ago

Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who

padilas [110]

Answer:

There is a 38.97% probability that this student earned an A on the midterm.

Step-by-step explanation:

The first step is that we have to find the percentage of students who got an A on the final exam.

Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.

This means that

Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.

So, the percentage of students who got an A on the final exam is

$P_{A} = 0.13(0.47) + 0.87(0.11) = 0.1568$

To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.

The percentage of students who got an A on both tests is:

$P_{AA} = 0.13(0.47) = 0.0611$

The probability that the student also earned an A on the midterm is

$P = \frac{P_{AA}}{P_{A}} = \frac{0.0611}{0.1568} = 0.3897$

There is a 38.97% probability that this student earned an A on the midterm.

5 0

1 year ago