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masha68 [24]
2 years ago
7

Which point (x,y) is a solution to the system of linear equations?

Mathematics
1 answer:
alukav5142 [94]2 years ago
5 0
You need to post the two equations that make up the “system”
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What is the equation of the line shown in this graph
prisoha [69]
Slope = (2+4)/(1-4) = 6/-3 = -2
passing thru (1.2)
y = mx + b
b = y - mx
b = 2 - (-2)(1)
b = 2 + 2
b = 4

equation
y = -2x + 4
7 0
2 years ago
Weights of cars passing over a bridge part 2<br>​
DaniilM [7]

Step-by-step explanation:

The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.

Given:

mean,  mu = 3550 lbs (hope I read the first five correctly, and it's not a six)

standard deviation, sigma = 870 lbs

weights are normally distributed, and assume large samples.

Probability to be estimated between W1=2800 and W2=4500 lbs.

Solution:

We calculate Z-scores for each of the limits in order to estimate probabilities from tables.

For W1 (lower limit),

Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069

From tables, P(Z<Z1) = 0.194325

For W2 (upper limit):

Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954

From tables, P(Z<Z2) = 0.862573

Therefore probability that weight is between W1 and W2 is

P( W1 < W < W2 )

= P(Z1 < Z < Z2)

= P(Z<Z2) - P(Z<Z1)

= 0.862573 - 0.194325

= 0.668248

= 0.67 (to the hundredth)

6 0
3 years ago
Solve the inequality 5 + 1/x &gt; 16/x
sladkih [1.3K]

Answer:

x  <  0  or     x  >  3

Explanation:

6 0
2 years ago
Read 2 more answers
In a random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand. C
horrorfan [7]

Answer:

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

Step-by-step explanation:

<em>Step(i)</em>:-

<em>Given sample size 'n' =300</em>

Given data  random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.

<em>Given sample proportion </em>

<em>                        </em>p^{-}  = \frac{x}{n} = \frac{182}{300} =0.606

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

<em>90% confidence interval for the proportion is determined by</em>

(p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} }  , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })

(0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} }  ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })

(0.6066 -  0.0463  ,0.6066 +  0.0463)

(0.5603,0.6529)

<u>final answer</u>:-

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

5 0
3 years ago
Can someone help me in stuck thank you (picture attached)
Ivahew [28]
C. 12 /3 go Brainly! Have Fun
5 0
3 years ago
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