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2 years ago

Can you please help answer the question in the photo?

Mathematics

1 answer:

Maurinko [17]2 years ago

7 0

Answer:

A: y= -x

B: y= 3x + 2

C: y= x + 2

D: y= x

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How would you estimate 412÷97?

marissa [1.9K]

Your answer would be 4.23 as the answer will be a long one, estimated.

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2 years ago

Where would you place .33 on the Venn diagram?

Mice21 [21]

it is a rational number (intergers are plus and minus whole numbers)

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3 years ago

A and B are independent events. If P(A) = 0.50 and P(A and B) = 0.20, find P(B).

rewona [7]

The independent events are illustrations of probability, and the value of P(B) is 0.40

<h3>How to determine the value of P(B)?</h3>

The given parameters are:

P(A) = 0.50

P(A and B) = 0.20

Two events A and B are independent, if

P(A and B) = P(A) * P(B)

So, we have:

0.50 * P(B) = 0.20

Divide both sides by 0.50

P(B) = 0.40

Hence, the value of P(B) is 0.40

Read more about probability at:

brainly.com/question/25638875

#SPJ1

5 0

1 year 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

6 0

3 years ago

230=4p+m solve for m

Sladkaya [172]

We can't find the numerical term for m but in terms of p it is possible.

230 = 4p + m
m = 230 - 4p

Hope this helps!

4 0

3 years ago

Read 2 more answers