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slamgirl [31]
3 years ago
6

Miss James has a six square-foot bulletin board in at 12 ft.² bulletin board she wants to cover both boards with index cards wit

hout gaps or overlaps each index card has an area of 1/4 ft.² how many index cards does she need
Mathematics
1 answer:
irinina [24]3 years ago
5 0
6 divided by 1/4 = 24 + 12 divided by 1/4 = 48, then 48 + 24 = 72
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Find the value of sin(a/2) if cosa= 12/13<br>​
daser333 [38]

Answer: sin\frac{a}{2} = ± \frac{1}{\sqrt{26} }

Step-by-step explanation:

We very well know that,

cos2A=1−2sin²A

⟹ sinA = ±\sqrt{(1-} \frac{cos2A}{2} )

As required,  set A = \frac{a}{2}   &   cos a=  \frac{12}{13}    ,thus we get

sin \frac{a}{2} =± \sqrt{\frac{1-cos a}{2} }  

∴ sin\frac{a}{2} =±\sqrt{\frac{1-\frac{12}{13} }{2} } = ± \frac{1}{\sqrt{26} }

   since ,360° < \frac{a}{2} <450°

             ,180° < \frac{a}{2} <225°

Now, we are to select the value with the correct sign. It's is obvious from the above constraints that the angle a/2 lies in the III-quadrant where 'sine' has negative value, thus the required value is negative.

hope it helped!

   

5 0
3 years ago
Help with this please
Gwar [14]

Answer:1

Step-by-step explanation:

4 0
3 years ago
What is the equation of a line that models wages with a slope of 15 that passes through the point (5, 90)?
Klio2033 [76]
Hello : 
<span>the equation is : 
y - 90 = 15(x - 5)</span>
3 0
3 years ago
The proportion of brown M&amp;M's in a milk chocolate packet is approximately 14%. Suppose a package of M&amp;M's typically cont
marshall27 [118]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the scenario above :

A) State the random variable.

The random variable is the p opoertion of brown M&M's in a milk chocolate packet.

B.) Argue that this is a binomial experiment.

Each trial is independent for a total number of 52 trials with a set probability of success at 0.14

C) probability that 6 M&M's are brown:.

P(x) = nCx * p^x * (1-p)^(n-x)

p = 0.14 ; (1 - p) = 0.86 ; n = 52 ; x = 6

P(x = 6) = 52C6 × 0.14^6 × 0.86^46

= 20358520 × 0.00000752954 × 0.00097035078

= 0.1487

D) P(x =25)

P(x = 25) = 52C25 × 0.14^25 × 0.86^27

= 477551179875952 × 449.987958058*10^(-24) × 0.01703955245

= 0.00000000366

E) P(x = 52)

P(x = 52) = 52C52 × 0.14^52 × 0.86^0

= 1 × 3968.78758299*10^(-48) × 1

= 3968.78758299*10^(-48)

F) yes it would be unusual, because such probability is extremely low. However, if a huge or substantial number of trials such may occur

3 0
3 years ago
HA=2-bA , solve the equation for A
yaroslaw [1]
HA + bA = 2
A(h+b) = 2
A = 2/h+b
7 0
2 years ago
Read 2 more answers
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