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2 years ago

9

Hiiiooos! Can someone please help me with this!!!❤️❤️❤️❤️:)

2 answers:

olga2289 [7]2 years ago

8 0

Answer: My best guess is 10 (sqrt of 2)

Step-by-step explanation:

Paha777 [63]2 years ago

5 0

Answer:

a

Step-by-step explanation:

hopes this helps

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What two consecutive integers have a sum of 75?

marshall27 [118]

Answer:

37 and 38.

Step-by-step explanation:

Let the smaller integer be x.

Thus, the larger integer is (x+1) since it is consecutive.

$x+(x+1)=75$

$2x+1=75$

Subtract both sides by 1:

$2x+1-1=75-1$

$2x=74$

Divide both sides by 2:

$\frac{2x}{2}=\frac{74}{2}$

$x=37$

To find the other number, simply add 1 to 37 to get 38.

7 0

3 years ago

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Find the rate of change from the given table.

Ymorist [56]

The rate of change would be D. 4

4 0

2 years ago

Evaluate-2+7x^2-2x^3-8x+7x^4 at x=5

Murljashka [212]

Answer:

4262

Step-by-step explanation:

2 + 7x^2- 2x^3-8x +7x^4 at x = 5

= 2 + 7*(5)^2 - 2*(5)^3-8*5 +7*(5)^4

=2+175-250-40+4375

=4552-290

=4262

4 0

3 years ago

What is 3÷4? Draw a diagram that explains how you know.

Yanka [14]

Answer:

3 divided by 4 is 0.75

Step-by-step explanation:

4 0

2 years ago

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Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

6 0

2 years ago