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11Alexandr11 [23.1K]
1 year ago
8

Explain why you would make one of the addends a tens number when solving an addition problem

Mathematics
1 answer:
leonid [27]1 year ago
4 0
<span> You refer to a multiple of ten. As a multiple of ten you can just add it to another number easier than a "ones number".</span>
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Use this information to find the value of b.<br><br><br>c = 25<br><br>s = 9<br><br><br>b = 4c - s2
Travka [436]
Hello There!

Write out the formula again:
b = 4c - s².
Substitute your numbers in:
b = 4(25) - (9)²
Simplify:
4 x 25 = 100
9² = 81
b = 100 - 81
Solve:
100 - 81 = 19
b = 19.

Hope This Helps You!
<span>Good Luck :) </span>

<span>- Hannah ❤</span>
3 0
1 year ago
Eight less than four times a number is less than 56. What are the possible values of that number
yan [13]
Let unknown number be x.

Given,

4x - 8 < 56

4x < 64

x < 16

Hence, x is any value less than 16.
4 0
1 year ago
Read 2 more answers
Each day, X arrives at point A between 8:00 and 9:00 a.m., his times of arrival being uniformly distributed. Y arrives independe
astraxan [27]

Answer:

Y will arrive earlier than X one fourth of times.

Step-by-step explanation:

To solve this, we might notice that given that both events are independent of each other, the joint probability density function is the product of X and Y's probability density functions. For an uniformly distributed density function, we have that:

f_X(x) = \frac{1}{L}

Where L stands for the length of the interval over which the variable is distributed.

Now, as  X is distributed over a 1 hour interval, and Y is distributed over a 0.5 hour interval, we have:

f_X(x) = 1\\\\f_Y(y)=2.

Now, the probability of an event is equal to the integral of the density probability function:

\iint_A f_{X,Y} (x,y) dx\, dy

Where A is the in which the event happens, in this case, the region in which Y<X (Y arrives before X)

It's useful to draw a diagram here, I have attached one in which you can see the integration region.

You can see there a box, that represents all possible outcomes for Y and X. There's a diagonal coming from the box's upper right corner, that diagonal represents the cases in which both X and Y arrive at the same time, under that line we have that Y arrives before X, that is our integration region.

Let's set up the integration:

\iint_A f_{X,Y} (x,y) dx\, dy\\\\\iint_A f_{X} (x) \, f_{Y} (y) dx\, dy\\\\2 \iint_A  dx\, dy

We have used here both the independence of the events and the uniformity of distributions, we take the 2 out because it's just a constant and now we just need to integrate. But the function we are integrating is just a 1! So we can take the integral as just the area of the integration region. From the diagram we can see that the region is a triangle of height 0.5 and base 0.5. thus the integral becomes:

2 \iint_A  dx\, dy= 2 \times \frac{0.5 \times 0.5 }{2} \\\\2 \iint_A  dx\, dy= \frac{1}{4}

That means that one in four times Y will arrive earlier than X. This result can also be seen clearly on the diagram, where we can see that the triangle is a fourth of the rectangle.

6 0
1 year ago
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 449 gram setting. It is
stealth61 [152]

Answer:

We accetp  H₀

Step-by-step explanation:

Information:

Normal distribution  

Population mean      =   μ₀  = 449

Population standard deviation  σ   unknown

Sample size   n  =  23        n < 30    we use t-student test

so   n  =  23    degree of fredom   df = n  - 1  df  = 23- 1   df = 22

Sample mean    μ =  448

Sample standard deviation   s  =  20

Significance level  α  =  0,05  

1.-Hypothesis Test

Null hypothesis                               H₀     μ₀  =  449

Alternative hypothesis                    Hₐ     μ₀  ≠  449

Problem statement ask for determine decision rule for rejecting the null hypothesis. For rejecting the null hypothesis we have to  get an statistic parameter wich implies  that μ is bigger or smaller than μ₀

2.-Significance level   α  =  0,05  ;  as we have a two tail test

α/2    =  0,025

Then from t - student table for  df =  22   and 0,025 (two tail-test)

t(c)  =  ±  2.074

3.- Compute  t(s)

t(s)   =  (  μ  -  μ₀ )  /  s /√n

plugging in values

t(s)   =  (448  -  449) /  20 /√23    ⇒   t(s)   =  -  1*√23 /20

t(s)   =  - 0.2398

4.-Compare t(c)   and  t(s)

t(s)  <  t(c)         - 0.2398  <  - 2.074

Therefore  t(s)  in inside acceptance region.  We accept  H₀

7 0
2 years ago
The formula that relates the length of a ladder, L, that leans against a wall with distance d from the base of the wall and the
Ulleksa [173]

Answer:

Step-by-step explanation:

The formula that relates the length of a ladder, L, that leans against a wall with distance d from the base of the wall and the height h that the ladder reaches up the wall is L = StartRoot d squared + h squared EndRoot. What height on the wall will a 15-foot ladder reach if it is placed 3.5 feet from the base of a wall?

L = √d² + h²

4 0
1 year ago
Read 2 more answers
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