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Lorico [155]
3 years ago
8

CEOLATIONS AND INEQUALITIES

Mathematics
1 answer:
lara [203]3 years ago
8 0

Answer:

seventeen degrees(17⁰)

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-6 &gt; -9<br><br> Which best explains whether the inequality is true or false?
Natali [406]

Answer:

-6 on a number line is closer to 0 which make sit greater while -9 is farther away which makes it less than -6

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Please help me solve 4 inequalities and show work. Will get brainiest!!
FrozenT [24]
1. y<span> ≤ 4x/3+5
2. y</span><span>< 6x/4+3</span>
5 0
3 years ago
Find x !!?????!!!????????????
mars1129 [50]

Answer:

x = 15

Step-by-step explanation:

Given

See attachment

Required

Find x

The figure in the attachment is a quadrilateral and the angles in a quadrilateral add up to 360.

So, we have:

90 + 6x + 5+ 10x - 40 + 4x + 5 = 360

Collect like terms

6x + 10x + 4x = 360 - 90 - 5 + 40 - 5

20x = 300

Divide both sides by 20

x = 15

Hence, the value of x is 15

4 0
2 years ago
Hiii! can someone please help me with my maths question
beks73 [17]

Step-by-step explanation:

a^2 + b^2 = c^2

(x+2)^2 + 4x^2 = (3x+4)^2

(x+2)(x+2)+ 4x^2= (3x+4)(3x+4)

x^2 + 2x + 2x + 4 + 4x^2 = 9x^2 + 12x + 12 x + 16

5x^2 + 4x + 4 = 9x^2 + 24x + 16

-4x^2 - 20x - 12 = 0

now we can use the quadratic formula

(-b±√(b²-4ac))/(2a)

I can't type all this so

.....

x = -0.697224362 or -4.302775638

AB can't be negative so it's the first one

so it's equals to 1.302775638

6 0
3 years ago
An instructor who taught two sections of engineering probability last term, the first with 20 students and thesecond with 30, de
Oliga [24]

Answer:

0.207

Step-by-step explanation:

This is an hypergeometric distribution problem

An hypergeometric distribution has the same sense as the discrete probabilities of binomial distribution, but unlike binomial distribution, hypergeometric distribution does not allow replacement.

Binomial distribution expresses the probability of picking k objects from n with replacement, but hypergeometric distribution expresses picking k objects from n without replacement, with the finite total population, N, containing K objects.

It is expressed mathematically as

h(k: n, K, N) = (ᴷCₖ)(ᴺ⁻ᴷCₙ₋ₖ)/(ᴺCₙ)

where

k = number of students in the 2nd section required to be in the first 15 graded projects (number of successes) = 10

n = total number of first graded projects (number of trials) = 15

K = number of students in the 2nd section of the class = 30

N = total number of students = 50

h(10: 15, 30, 50) = (³⁰C₁₀)(⁵⁰⁻³⁰C₁₅₋₁₀)/(⁵⁰C₁₅)

h(10: 15, 30, 50) = (³⁰C₁₀)(²⁰C₅)/(⁵⁰C₁₅)

= (30,045,015)(15,504)/(2,250,829,575,120)

P(X = 10) = 0.207

Hope this Helps!!!

7 0
3 years ago
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