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3 years ago

10

The table below shows the cost for a factory to produce mid-sized cars.

1 answer:

quester [9]3 years ago

6 0

Just divide them

cost/3cars=16194/3cars=5398/1car

answer is A

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Which must be true?

Eva8 [605]

Answer:

PSC=PTC

They are the same angle

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3 years ago

Angle Measurements

Naddik [55]

Answer:

X = 11

Complementary angles

Step-by-step explanation:

Complementary angles are when two angles add up to 90 degrees.

3x + 10 + 47 = 90

3x + 10 = 43

3x = 33

x = 11

4 0

3 years ago

Please help me with these, oh sweet jesus

Lelechka [254]

Answer:

77. $\cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} x$ Proved

78. $\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]$ Proved

79. $\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x$ Proved.

80. $\sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x$ Proved.

Step-by-step explanation:

77. Left hand side

= $\cot^{6} x$

= $\cot^{4} x \times \cot^{2} x$

= $\cot^{4}x [\csc^{2}x - 1]$

{Since we know, $\csc^{2} x - \cot^{2}x = 1$ }

= $\cot^{4} x \csc^{2}x - \cot^{4} x$

= Right hand side (Proved)

78. Left hand side

= $\sec^{4}x \tan^{2} x$

= $\sec^{2} x [1 + \tan^{2}x] \tan^{2} x$

{Since $\sec^{2}x - \tan^{2}x = 1$ }

= $\sec^{2}x [\tan^{2}x + \tan^{4}x ]$

= Right hand side (Proved)

79. Left hand side

= $\cos^{3} x\sin^{2} x$

= $\cos x[1 - \sin^{2} x] \sin^{2} x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $[\sin^{2}x - \sin^{4}x] \cos x$

= Right hand side

80. Left hand side

= $\sin^{4}x - \cos^{4}x$

= $[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $1 - 2\cos^{2} x[1 - \cos^{2}x ]$

= $1 - 2\cos^{2}x + 2 \cos^{4} x$

= Right hand side. (Proved)

7 0

3 years ago

What are the integer solutions to the inequality below?<br> <img src="https://tex.z-dn.net/?f=3%5Cleq%203x-4%5C%20%20%5Ctextless

Rudik [331]

Given:

The compound inequality is:

$3\leq 3x-4$

To find:

The integer solutions for the given compound inequality.

Solution:

We have,

$3\leq 3x-4$

Case 1: $3\leq 3x-4$

$3+4\leq 3x$

$\dfrac{7}{3}\leq x$

$2.33...\leq x$ ...(i)

Case 2: $3x-4$

$3x-2x$

$x$ ...(ii)

Using (i) and (ii), we get

$2.33...$

The integer values which satisfy this inequality are only 3 and 4.

Therefore, the integer solutions to the given inequality are 3 and 4.

7 0

2 years ago

In Illinois, 9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders; that is, they have been arre

Veseljchak [2.6K]

Answer:

a) 21.58% probability that exactly 3 people are repeat offenders

b) 97.91% probability that at least one person is a repeat offender

c) 3.69

d) 1.83

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders

This means that $p = 0.09$

41 people arrested for DUI in Illinois are selected at random.

This means that $n = 41$

a. What is the probability that exactly 3 people are repeat offenders?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{41,3}.(0.09)^{3}.(0.91)^{38} = 0.2158$

21.58% probability that exactly 3 people are repeat offenders

b. What is the probability that at least one person is a repeat offender?

Either none are repeat offenders, or at least one is. The sum of the probabilities of these outcomes is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ .

Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = C_{41,0}.(0.09)^{0}.(0.91)^{41} = 0.0209$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0209 = 0.9791$

97.91% probability that at least one person is a repeat offender

c. What is the mean number of repeat offenders?

$E(X) = np = 41*0.09 = 3.69$

d. What is the standard deviation of the number of repeat offenders?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{41*0.09*0.91} = 1.83$

4 0

2 years ago