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3 years ago

Y= 5x -3 linear relation missing values

Mathematics

2 answers:

Alja [10]3 years ago

7 0

Answer:

when x=-2,y=-13

when y=-52,x=-9.8

when x=0,y=-3

when y=0,x=3/5 or 0.6

hope it helps!!!!!

Nimfa-mama [501]3 years ago

7 0

Y=5x+3

let x=0. 5(0)+3=3 Hence this linear function value of three at x=0

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What is the solution for x in the equation

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a sprinkler that sprays water in a circular area can be adjusted to spray up to 30 feet. turning the radius-reduction screw on t

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3 years ago

Read 2 more answers

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Answer:

$1.\ \ np\geq 10\\\\2.\ \ n\geq 10x$

Step-by-step explanation:

-Let x be the sample size and n the population size

-The conditions for a one-sample proportion z-test are:

-The sample is randomly selected from the population.

-The sample size is greater than or equal to ten times the population size:

$n\geq 10x\\\\n=2000\\x=150\\10x=1500\\\therefore n\geq 10x$

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4 years ago

A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a

natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student

The sample means = x₁ + x₂ + . . . , x₂₃ / 23

$= \frac{x_1 +x_2 +...x_2_3}{23}$

= 2.3hr

⇒ $\sum xi =2.3 \times 23 = 52.9hrs$

let x₁ , x₂, . . . , x₂₃ arranged in ascending order

Then x₂₃ was 10 and has been changed to 14

i.e x₂₃ increase to 4

Sample mean $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)² ∑ (xi - 2.47)²

Therefore , the standard deviation increases

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4 years ago