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2 years ago

Can someone please answer for d and b

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

3 0

Answer:

d=2√2, b= $\frac{3\sqrt{2}}{2}$

Step-by-step explanation:

first triangle is a 30-60-90

so, sides x-x√2-2x

x=2

therefore d=2√2

second triangle is 45-45-60

sides, 1-1-√2

which means b√2=6

b=6/√2 ---> multiply by √2/√2 to get ride of the √2 denominator

b=6√2/4 , simplify

b= $\frac{3\sqrt{2}}{2}$

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Among 500 freshmen pursuing a business degree at a university, 315 are enrolled in an economics course, 213 are enrolled in a ma

scoundrel [369]

Answer:

Step-by-step explanation:

Given that among 500 freshmen pursuing a business degree at a university, 315 are enrolled in an economics course, 213 are enrolled in a mathematics course, and 123 are enrolled in both an economics and a mathematics course.

From the above we find that

a) either economics of Math course is

$315+213-123 =405$

Out of 500 students 405 have taken either Math or Economics

Hence

c) student who have taken neither = $500-405 =95$

Exactly one course is either math or economics - both

= $405-123 = 282$

3 0

3 years ago

The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that,

professor190 [17]

Answer:

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Mean of 0.6 times a day

7 day week, so $\mu = 7*0.6 = 4.2$

What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103$

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

3 0

3 years ago

one rectangle measures 2 units by 7 units. a second rectangle measures 11 units by 37 units. are these two figures scaled versio

Amanda [17]

Answer: The both triangle are not scale version to each other.

Explanation:

It is given that the one rectangle measures 2 units by 7 units. a second rectangle measures 11 units by 37 units.

The two shapes are scaled version of each other if their corresponding sides are in same proportion. In other words the if two shapes scaled version of each then the shapes are similar.

The sides or first rectangle are 2 and 7. The measures of second rectangle are 11 and 37.

The proportion of small side is,

$\frac{2}{11}$

The proportion of large side is,

$\frac{7}{37}$

So,

$\frac{2}{11}\neq \frac{7}{37}$

Since the proportion are not equal, therefore both triangle are not scale version to each other.

6 0

3 years ago

-4/3(3q-10) simplified

monitta

Answer:

Answer in the photo

Step-by-step explanation:

Just simplify the expression

4 0

3 years ago

Read 2 more answers

Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.

masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is $g^{-1}(x) = e^{x}$

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is $g^{-1}(x) = e^{x}$

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y = ln(x)

Taking exponents of both sides, we have

$e^{y} = e^{lnx} \\e^{y} = x$

Replacing x with y, we have

$y = e^{x}$

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is $g^{-1}(x) = e^{x}$

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

5 0

2 years ago