All categories

3 years ago

Will the lengths 4, 9, 5 form a triangle

Mathematics

1 answer:

seropon [69]3 years ago

6 0

Answer:

no.

Step-by-step explanation:

The sum of two side lengths have to be greater than the value of the remaining side length.

Hope this helps. Have a nice day you amazing bean child.

You might be interested in

Can someone help me out please??

olganol [36]

Answer:

the 2nd one is the answer.

x ≥ −8

Step-by-step explanation:

Hope that helps

Sorry if wrong

8 0

3 years ago

Help I’m tired it’s 2:56am it’s due tomorrow morning !

guajiro [1.7K]

Answer:

go to sleep just take the 0 and thinks for the points

Step-by-step explanation:

8 0

2 years ago

If the

ahrayia [7]

It’s A bc it equals to 8

8 0

3 years ago

(1) The ratios 18:3 and 6:1 are equivalent. Why?

aliya0001 [1]

Answer:

Because if you take 18:3 and divide both sides by 3, you will get 6:1

Step-by-step explanation:

Because 18:3 divided by 3 on both sides equals 6:1, also 6:1 times 3 on both sides equals 18:3.

6 0

3 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

6 0

3 years ago