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Bad White [126]

3 years ago

9

Lani had some flowers in her garden from last year. When spring arrived, Lani started planting more flowers in the garden each w

eek. The total
numbers of flowers in the garden after weeks 2, 4, 6, and 8 were 14, 18, 22, and 26, respectively. How many flowers did Lani have in her garden
from last year? Assume that the relationship between the two quantities is linear.
a.2
b.4
c.10
d.12

1 answer:

AURORKA [14]3 years ago

4 0

It's a.2 because it is

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The temperature at noon is -4 degrees Fahrenheit. The temperature at 6:00pm is -12 degrees fahrenheit. what is the difference be

klio [65]

Answer:

8 degrees difference

Step-by-step explanation:

Just subtract the two numbers given

-4--12

-4+12=8

(not -8 because it’s the difference between the NOON and the 6pm tempature)

6 0

3 years ago

Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of

Andrews [41]

Answer:

$t = 31.29$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{Stick} & {25.8} & {26.9} & {26.2} & {25.3} & {26.7}& {26.1} \ \\ {Liquid} & {16.9} & {17.4} & {16.8} & {16.2} & {17.3}& {16.8} \ \end{array}$

Required

Determine the test statistic

Let the dataset of stick be A and Liquid be B.

We start by calculating the mean of each dataset;

$\bar x =\frac{\sum x}{n}$

n, in both datasets in 6

For A

$\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}$

$\bar x_A =\frac{157}{6}$

$\bar x_A =26.17$

For B

$\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}$

$\bar x_B =\frac{101.4}{6}$

$\bar x_B =16.9$

Next, calculate the sample standard deviation

This is calculated using:

$s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

For A

$s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}$

$s_A = \sqrt{\frac{(25.8-26.17)^2+(26.9-26.17)^2+(26.2-26.17)^2+(25.3-26.17)^2+(26.7-26.17)^2+(26.1-26.17)^2}{6-1}}$

$s_A = \sqrt{\frac{1.7134}{5}}$

$s_A = \sqrt{0.34268}$

$s_A = 0.5854$

For B

$s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}$

$s_B = \sqrt{\frac{(16.9 - 16.9)^2+(17.4- 16.9)^2+(16.8- 16.9)^2+(16.2- 16.9)^2+(17.3- 16.9)^2+(16.8- 16.9)^2}{6-1}}$

$s_B = \sqrt{\frac{0.92}{5}}$

$s_B = \sqrt{0.184}$

$s_B = 0.4290$

Calculate the pooled variance

$S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}$

$S_p^2 = \frac{(6 - 1)*0.5854^2 + (6 - 1)*0.4290^2}{(6+6-2)}$

$S_p^2 = \frac{2.6336708}{10}$

$S_p^2 = 0.2634$

Lastly, calculate the test statistic using:

$t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_B)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

We set

$\mu_A = \mu_B$

So, we have:

$t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_A)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

$t = \frac{(\bar x_A - \bar x_B) }{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

The equation becomes

$t = \frac{(26.17 - 16.9) }{\sqrt{0.2634/6 +0.2634/6}}$

$t = \frac{9.27}{\sqrt{0.0878}}$

$t = \frac{9.27}{0.2963}$

$t = 31.29$

<em>The test statistic is 31.29</em>

3 0

2 years ago

Is 6/10 larger than 9/15?

alexandr1967 [171]

6/10 is larger than 9/15's, good luck on your test!

6 0

3 years ago

If a = 5 in the right triangle what is the value of b and c

Nadya [2.5K]

Answer:

b=5√3, c=10

Step-by-step explanation:

30-60-90 triangle

c=2a

b=a√3

6 0

3 years ago

What is the measure of Arc BA? An explanation would be much appreciated.

yKpoI14uk [10]

Answer:

m<BA = 180degrees

Step-by-step explanation:

First you must know that measure of <BA is 180 degrees and m<CD = m<BA

Since <CD = 16z - 12, hence;

180 = 16z - 12

16z = 180++12

16z = 192

z = 192/16

z = 12

Get <BA

m<BA = 16z - 12

m<BA = 16(12) - 12

m<BA = 192 - 12

m<BA = 180degrees

8 0

3 years ago