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mezya [45]
3 years ago
7

Compute.

Mathematics
1 answer:
Sholpan [36]3 years ago
8 0
<span>6 yr 7 months+ 2 yr 11 months = ?
6 yr + 2 yr = 8 yr

7 months + 11 months = 18 months = 1 yr 6 months

8 yr + </span>1 yr 6 months = 9 yr 6 months

answer is C<span>. 9 yr 6 months</span>
You might be interested in
What is the radian measure of the smaller angle formed by the hands of a clock at 4 pm?
never [62]
On a clock, every 5 minutes is just another 30° angle deviation away from the 12th hour, hence 30° × 4 = 120°

Converting this value into radians..

\frac{120\pi}{180}
Simplifying this value should get you, 2π/3 rad, which is the radian measure of the clock hands at 4pm.

4 0
3 years ago
How do I solve this problem? ​
natulia [17]

Step-by-step explanation:

you know the formulas ?

the area of a circle is : pi×r²

the circumference of a circle is : 2×pi×r

with r being the radius, which is half of the diameter.

so, here, r = 10/2 = 5 m

area = pi×r² = pi×5² = pi×25 or 25×pi m²

circumference = 2×pi×r = 2×pi×5 = 10×pi m

for the approximation we make the calculations with pi on the calculator and round to the nearest hundredth.

area = 25×pi ≈ 78.54 m²

circumference = 10×pi ≈ 31.42 m

5 0
3 years ago
PLEASE HELP ME!!! Which expression is equivelent to 4(9+7) A. 36 + 7 B. 13 + 11 C. 36 + 28 D. 13 + 7
aleksandrvk [35]
Using the distributive property , 4*9+4*7 or 36+28
6 0
3 years ago
Read 2 more answers
If jobs arrive every 15 seconds on average, what is the probability of waiting more than 30 seconds?
Aleksandr [31]

Answer: 0.14

Step-by-step explanation:

Given: Mean : \lambda=15\text{ per seconds}

In minutes , Mean : \lambda=4\text{ per minute}

The exponential distribution function with parameter \lambda  is given by :-

f(t)=\lambda e^{-\lambda t}, \text{ for }x\geq0

The probability of waiting more than 30 seconds i.e. 0.5 minutes is given by the exponential function :-

P(X\geq0.5)=1-P(X\leq0.5)\\\\=1-\int^{0.5}_{0}4e^{-4t}dt\\\\=1-[-e^{-4t}]^{0.5}_{0}\\\\=1-(1-e^{-2})=1-0.86=0.14

Hence, the probability of waiting more than 30 seconds = 0.14

5 0
3 years ago
Find the x-intercepts of the parabola with vertex (1,1) and y-intercept (0,-3). Write your answer in this form (x of 1 , y of 1)
Vikki [24]
Remember that the vertex form of a parabola or quadratic equation is:

y=a(x-h)^2+k, where (h,k) is the "vertex" which is the maximum or minimum point of the parabola  (and a is half the acceleration of the of the function, but that is maybe too much :P)

In this case we are given that the vertex is (1,1) so we have:

y=a(x-1)^2+1, and then we are told that there is a point (0,-3) so we can say:

-3=a(0-1)^2+1

-3=a+1

-4=a so our complete equation in vertex form is:

y=-4(x-1)^2+1

Now you wish to know where the x-intercepts are.  x-intercepts are when the graph touches the x-axis, ie, when y=0 so

0=-4(x-1)^2+1  add 4(x-1)^2 to both sides

4(x-1)^2=1  divide both sides by 4

(x-1)^2=1/4  take the square root of both sides

x-1=±√(1/4)  which is equal to

x-1=±1/2  add 1 to both sides

x=1±1/2

So x=0.5 and 1.5, thus the x-intercept points are:

(0.5, 0) and (1.5, 0)  or if you like fractions:

(1/2, 0) and (3/2, 0) :P
4 0
3 years ago
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