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3 years ago

The unit digit of cube 149 is

Mathematics

2 answers:

Eva8 [605]3 years ago

8 0

Unit digit of cube 149 is 7.

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ICE Princess25 [194]3 years ago

4 0

The unit digit of cube 149 is 9

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Tripp and Rico are two dogs. Tripp weighs exactly 35 pounds more than Rico. Together, they weigh exactly 49 pounds. How much doe

babymother [125]

Call the weights of the two dogs $t$ and $r$ .

We know that Tripp weights 35lbs more than Rico, or:
$t=r+35$

We also know that the total weight of both dogs is 49lbs.:
$t+r=49$

Now, by substitution:
$(r+35)+r=49$
$2r+35=49$
$2r+35-35=49-35$
$2r=14$
$r=\frac{14}{2}$
$r=7$
Rico weighs 7lbs.

Then:
$t=r+35$
$t=7+35$
$t=42$
Tripp weighs 42lbs.

To check our work:
$t+r=49$
$42+7=49$
$49=49$ CHECK!

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3 years ago

Please help!! Problem in picture

notsponge [240]

20.if m <6=72, then m < 7=108
21. if m<8,then m<7=80
22.if m<110,then m<6=90
23.if m<123, then m<8=123
24.if m<142, then m<7=38
25.if m<13, then m<8=167
26.if m<170, then m<10
27.if m<26, then m <154
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3 years ago

Four int variables, x1, x2, y1, and y2, have been declared and been given values. write an expression whose value is the differe

Katen [24]

Even though we are using variables, we still know that "difference" means subtraction. So, if the larger is x1 and x2, those variables go first. The smaller, y1 and y2 will go second. So our problem will look like this:

(x1 < x2 ? x2 : x1) - (y1 < y2 ? y1 : y2)

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3 years ago

This is so confusing please help asap !

Tju [1.3M]

I don’t know how sorta

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2 years ago

Read 2 more answers

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.

(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:

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