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Lisa [10]
3 years ago
14

Write an algebraic equation to represent the following problem:

Mathematics
1 answer:
Elena L [17]3 years ago
4 0

Answer:

$47.60

Step-by-step explanation:

18+\frac{148}{5} =47.60 ($)

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On a sheet of paper, you have 100 statements written down.
serg [7]

Answer:

50

Step-by-step explanation:

If the first statement its true (At most 0 of the statements are true), there are not true statements in the paper. So, the first statement its false.

Now, if the first statement its false, this mean there must be at least 1  true statement in the paper.

Now, if the second statement its true ( at most 1 of the statements are true) this implies that the third statement its true (if "at most 1" its true, then "as most 2" must be true).

If any statement (besides the first) its true, then all the statement that follows it must be true.

The first non false statement, then, must be the statement made by the person 51:  "At most 50 statements are true"

And the 49 statements that follows are true as well.

5 0
3 years ago
the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai
Blababa [14]

Let X be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10

where Z follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)

where F_Z(z)=P(Z\le z) denotes the CDF of Z, and F_Z^{-1} denotes the inverse CDF. We have

z_1=F_Z^{-1}(0.90)\approx1.2816

Similarly, because 20% of students obtain less than 40 marks, we have

P(X

so that

P(Z

Then \mu,\sigma are such that

\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma

\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma

and we find

\mu\approx53.8739,\sigma\approx16.4848

3 0
4 years ago
An object is launched from a platform. Its height (in meters), xxx seconds after the launch, is modeled by: h(x)=-5x^2+20x+60h(x
EastWind [94]
It takes 6 seconds for it to hit the ground.

0 = -5x²+20x+60

We can solve this by factoring.  First factor out the GCF, -5:

0 = -5(x²-4x-12)

Now we want factors of -12 that sum to -4.  -6(2) = -12 an -6+2 = -4:
0 = -5(x-6)(x+2)

Using the zero product property, we know that either x-6=0 or x+2=0; this gives us the answers x=6 or x=-2.  Since we cannot have negative time, x=6.
3 0
4 years ago
Read 2 more answers
Please help me with 13 and 15 asap
Law Incorporation [45]
Gosh, I've done this problem before. Let's start with 13. In this problem, we're basically just skip counting. For example, in the roses row, in the second bouquet, we know we have to add 4 more flowers, so we can document 8. Continue to skip count for both. For 15, we would have about 96 more movie posters remaining, making our ratio 96:x. So, 96:x = 120:100. Therefore, x would equal 80- as 96:80 equals 120:100. If she needs 80 and already had 100, she should sell 20 posters. Hope this helped.
3 0
3 years ago
Which of the following vectors are orthogonal to (1,5)? Check all that apply
Andre45 [30]
We know that
If the scalar product of two vectors<span> is zero, both vectors are </span><span>orthogonal

</span><span>A. (-2,5)
</span>(-2,5)*(1,5)-------> -2*1+5*5=23-----------> <span>are not orthogonal 

</span><span>B. (10,-2)
</span>(10,-2)*(1,5)-------> 10*1-2*5=0-----------> are  orthogonal 

<span>C. (-1,-5)
</span>(-1,-5)*(1,5)-------> -1*1-5*5=-26-----------> are not orthogonal 

<span>D. (-5,1)
</span>(-5,1)*(1,5)-------> -5*1+1*5=0-----------> are  orthogonal 

the answer is 
B. (10,-2) and D. (-5,1)  are  orthogonal  to  (1,5)
7 0
3 years ago
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