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Alexus [3.1K]
3 years ago
9

30 points! The solutions to the equation: 2x2 + x-1 = 2 are x=-3/2 or X=___

Mathematics
1 answer:
frez [133]3 years ago
3 0

Answer: A quadratic equation is one of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0. So I think x = 4 but not 100%

Sorry if I'm wrong but I hope this helps

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Kira drew PQR and STU so that P S, Q T, PR = 12, and SU = 3. Are PQR and STU similar? If so, identify the similarity postulate o
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3. Product of two numbers is 27/14. If one of them is 1 2/7, find the<br>other.<br>​
Tems11 [23]

Answer:

\frac{9}{8}

Step by step explanation:

\frac{27}{14}  \div  \frac{12}{7}  \\  \:  \:  \:   \:  \:  \:  \:  \:   \:  \:  \:  \: \:  =  \frac{27}{14}   \times  \frac{7}{12}  =  \frac{9}{8}

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3 years ago
What's the slope of the lines: (-9,5) and (-3,3)
Leokris [45]
The slope of the line is -2/6

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3 years ago
The distribution of the weights of loaves of bread from a certain bakery follows approximately a normal distribution. Based on a
artcher [175]

Answer:

μ=23.710740....

σ=3.592592....

Step-by-step explanation:

At first, we are told that it´s a normal distribution and the following:

P(x≤15.34)=0.1

P(x≤16.31)=0.2

We start looking for those values (0.1 and 0.2) and the probabilities that gives us those values in a normal standard distribution. There´s some tools that´ll help us out but i´ll be using a chart with the values of some probabilities in a normal standard distribution (That it´s attached).

The probability of P(z≤ a.b) where "a" is the whole part of a number and "b" the decimal part is in the coordinates (a,b). For example for our problem:

P(z≤-2.33)=0.099≈0.1

P(z≤-2.06)=0.197≈0.2

To find a probability of a normal (not standard) distribution we use a method called Normalize that proceeds:

P(x≤a)=P( (x-μ)/σ ≤ (a-μ)/σ )

Where μ is the Mean of the data and σ is the Standard deviation. P( (x-μ)/σ ≤ (a-μ)/σ ) is now part of a normal standard distribution and we are able to look for it in the chart.

We will use it to find the Mean and the STD for our distribution.

P(x≤15.34)=P(z≤-2.33)≈0.1

P(x≤16.31)=P(z≤-2.06)≈0.2

so we have the equations:

(15.34 - μ)/σ= -2.33 and (16.31 - μ)/σ= -2.06

And we solve them for μ and σ:

(15.34 - μ)/-2.33= σ and (16.31 - μ)/-2.06= σ

(15.34 - μ)/-2.33 = (16.31 - μ)/-2.06

2.06*(15.34 - μ) = 2.33(16.31 - μ)

2.33μ - 2.06μ = 2.33(16.31) - 2.06(15.34)

μ= 6.4019 / 0.27

μ= 23.710740...

With this value we find σ in one equation:

(15.34 - (23.710740...)/σ = -2.33

(15.34 - (23.710740...)/-2.33 = σ

σ=3.592592...

We finally have our answers:

μ=23.710740....

σ=3.592592....

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