For the displayed triangle, provide the following value B=

A pipe s leaking at 1.5 cups per day how many gallons per week is the pipe leaking

kkurt [141]

1.77 so i wish u good lock

8 0

3 years ago

What is b + 1.1 = - 11

gladu [14]

B + 1.1 = -11

Subtract 1.1 from both sides to get b.

So, b = -12.1

7 0

3 years ago

Read 2 more answers

2/8 + 6/12 what is that eqaul to

Papessa [141]

$\frac{2}{8} + \frac{6}{12} \\ \\ \frac{1}{4} + \frac{1}{2} \\ \\ LCD = 4 \\ \\ \frac{1}{4} + \frac{2}{4} \\ \\ \frac{1+2}{4} \\ \\ \frac{3}{4} \\ \\ Answer: \fbox {3/4} \ or \ \fbox {0.75}$

8 0

3 years ago

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Find the equation of the line passing through the points 1,-5) and (9,11). y = 2x + [?]

Naily [24]

Answer:

y = 2x - 7

Step-by-step explanation:

Looks like we already have the slope of this line: It is 2. Working with the point (1, -5), we have x = 1 and y = -5 and can from this info easily find the y-intercept, b:

y = mx + b becomes

y = 2x + b, which in turn becomes

-5 = 2(1) + b, or

b = -7,

and so the desired equation is y = 2x - 7

3 0

2 years ago

Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c

GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(c) The value of P (X < 30) is 0.9502.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

$f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0$

The value of E (X) is 10.

The parameter λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10$

(a)

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679$

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

$P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353$