Can someone pls pls pls pls pls pls answer these question

A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a

Diano4ka-milaya [45]

Answer: (C) 0.1591

Step-by-step explanation:

Given : A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a normal distribution with

$\mu=42,100\text{ miles}$

$\sigma=2,510\text{ miles}$

Let x be the random variable that represents the lifetime of the tires .

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x= 44,500 miles

$z=\dfrac{44500-42100}{2510}\approx0.96$

For x= 48,000 miles

$z=\dfrac{48000-42100}{2510}\approx2.35$

Using the standard normal distribution table , we have

The p-value : $P(44500$

$P(z$

Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles = 0.1591

3 0

3 years ago

Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required

xenn [34]

Answer:

(a) 8.15

(b) 12.92

Step-by-step explanation:

Given: P = $3000, r = 0.085

$A = Pe^{rt}$

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

A = 2 × $3000

A = $6000

$6000 = 3000e^{0.085t}$

$\frac{6000}{3000} = e^{0.085t}$

$2 = e^{0.085t}$

Take $log_{e}$ of both sides

$log_{e}2 = log_{e}e^{0.085t}$

But $log_{e}e = 1$

∴ $ln2 = 0.085t$

$t = \frac{ln2}{0.085}$

$t = \frac{0.693}{0.085}$

t = 8.15

(b) For the amount to double, A = 3 × P

A = 3 × $3000

A = $9000

$9000 = 3000e^{0.085t}$

$\frac{9000}{3000} = e^{0.085t}$

$3 = e^{0.085t}$

Take $log_{e}$ of both sides

$log_{e}3 = log_{e}e^{0.085t}$

But $log_{e}e = 1$

∴ $ln3 = 0.085t$

$t = \frac{ln3}{0.085}$

$t = \frac{1.0986}{0.085}$

t = 12.92

5 0

3 years ago

The graph of f ′ (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f

Nataliya [291]

The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.

Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.

By the fundamental theorem of calculus,

$\displaystyle f(5) = f(0) + \int_0^5 f'(x) \, dx$

The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so

$\displaystyle \int_0^5 f'(x) \, dx = \frac{5+2}2 \times 2 = 7$

$\implies \max\{f(x) \mid 0\le x \le5\} = f(5) = f(0) + 7 = \boxed{13}$

8 0

3 years ago

Answer the following questions about slope and y-intercept.

Juliette [100K]

Answer:

1. The y-intercept

2. The slope of the equation represent the relationship between the time duration in which Clarissa eats chocolate is 1/10

Step-by-step explanation:

1. The value of the y coordinate at the point which the graph meets the y-axis is called to the y-intercept. It is the value at which the x-coordinate is equal to zero, that is, the coordinate of the point at the y-intercept = (y, 0)

2. Given that Clarissa eats 1 piece of chocolate every 10 seconds, we have the slope of the equation represent the relationship between the time duration in which Clarissa eats chocolate is 1/10.

6 0

3 years ago

Evaluate: 6-(2/3)^2 A. 17/3 B. 52/3 C. 49/9 D. 50/9

Komok [63]

Answer:

The correct answer is: "Option [D]".

Step-by-step explanation:

Hi student, let me help you out!

....................................................................................................................................

Let's use the acronym PEMDAS. With the help of this little acronym, we will not make mistakes in the Order of Operations! :)

$\dag\textsf{Acronym \: PEMDAS}$

P=Parentheses,

E=Exponents,

M=Multiplication,

D=Division,

A=Addition,

S=Subtraction.

Now let's start evaluating our expression, which is $\mathsf{6-(\cfrac{2}{3})^2}$

According to PEMDAS, the operation that we should perform is "E-Exponents".

Notice that we have a fraction raised to a power. When this happens, we raise both the numerator (2 in this case) and the denominator (3 in this case) to that power, which is 2. After this we obtain $\mathsf{6-\cfrac{4}{9}}$ .

See, we raised both the numerator and the denominator to the power of 2.

Now what we should do is subtract fractions.

Note that 6 and -4/9 have unlike denominators. First, let's write 6 as a fraction: $\mathrm{\cfrac{6}{1}-\cfrac{4}{9}}$ . Now let's multiply the denominator and the numerator of the first fraction times 9: $\mathrm{\cfrac{54}{9}-\cfrac{4}{9}}$ .

See, now the fractions have the same denominator. All we should do now is subtract the numerators: $\mathrm{\cfrac{50}{9}}$ .

∴, the answer is Option D.

Hope this helped you out, ask in comments if any queries arise.

Best Wishes!

$\star\bigstar\underline{\overline{\overline{\underline{\textsf{Reach \: far. Aim \: high. Dream \: big.}}}}}\bigstar\star$

$\underline{\rule{300}{5}}$

5 0

2 years ago

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