YOOOOOOOOOOOOO HEELPPPPPPPPPPPPPPPP

As she drove down the icy Road Miss Campbell slammed on her breaks her car did a 360 explain what happened to her car

meriva

Answer:

The friction that the car usually uses to stop(a rough and hard ground) was null and void, so the car spun uncontrollably.

3 0

3 years ago

17. Tiana deposited $500 into an

Ivahew [28]

Answer:

In 50 years the amount will be $ 2000.

Step-by-step explanation:

Tiana deposited $500 into a account at 6 % simple interest per Year .

Therefore, Principal (P) is $500

Amount (Principal + Interest) = $2000

Interest (I) = $2000 - $500 = $1500

Rate percentage (R) = 6% per Year

Let, The time be T years

Now, I = $\frac{P . R . T}{100}$

Or, T = $\frac{100. I }{P . R}$

Or, T = (100×1500 ) ÷ ( 500× 6 )

<em> </em><em>∴ T = 50 Years</em>

3 0

3 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

(c) $P(X > 980) = 0.00001\\\\$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

3 years ago

Find the measure of angle A.

Ivanshal [37]

Answer:

∠A = 32°

Step-by-step explanation:

Using the tangent ratio in the right triangle

tanA = $\frac{opposite}{adjacent}$ = $\frac{5}{8}$ = 0.625, thus

A = $tan^{-1}$ (0.625) = 32°

4 0

3 years ago

Solve the equation for x:

never [62]

3x - 17 = 9x + 7

Rearrange on same sides
-17 - 7 = 9x - 3x
-24 = 6x

Divide by 6 on either sides to isolate 6
$-\frac{24}{6} = \frac{6x}{6}$
6 and 6 cancels out
-4 = x

Answer is:
B. x = -4

8 0

3 years ago