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2 years ago

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Natalia is writing a recursive formula to represent the sequence 8,12,18,27 what value should she use as the common ratio in the

formula? write the answer as a decimal rounded to the nearest tenth place

2 answers:

Marrrta [24]2 years ago

7 0

Common ratio = second term / first term = 12 / 8 = 1.5

BigorU [14]2 years ago

5 0

Answer:

The answer is 1.5.

Good luck and have a great day :)

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Which model shows 64−−√3=4 ?

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Yolanda used a reconciliation worksheet provided by her bank. Line 5 should equal line 6 at the end of the reconciliation, but i

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Answer is 19.99 dollars

Explanation:

Let us draw the reconciliatin done by Yolanda linewise

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3) Sum of 1 and 2 345.89

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SInce difference is exactly 19.99 dollars Yolanda should look for the amount 19.99 dollars to find a solution to the problem i.e. to see the reason for the difference between lines 5 and 6

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1 year ago

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Suppose a snow cone has a paper cone that is 8 centimeters deep and has a diameter of 5 centimeters. The flavored ice comes in a

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check the picture below.

so the cone itself has a height of 8, and a radius of 2.5, and the spherical scoop, ready to melt in it, has a radius of 2.5.

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$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=2.5\\ h=8 \end{cases}\implies V=\cfrac{\pi (2.5)^2(8)}{3}\implies \implies \boxed{V=\cfrac{50\pi }{3}} \\\\\\ \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}\qquad r=2.5\implies V=\cfrac{4(\pi )(2.5)^3}{3}\implies \boxed{V=\cfrac{125\pi }{6}} \\\\\\ \textit{the sphere's volume is larger, so yes, it will overflow}$

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Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

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1 year ago