Which figure shows 25% shaded?

Bernardo and Ogechi were asked to find an explicit formula for the sequence 1\,,\,8\,,\,64\,,\,512,...1,8,64,512,...1, comma, 8,

MatroZZZ [7]

Answer:

$h_{n}=1.(8)^{n-1}$ will be the correct formula for the given sequence.

Step-by-step explanation:

The given sequence is 1, 8, 64, 512...........

The given sequence is a geometric sequence having a common ratio (r) of

r = $\frac{\text{Second term}}{\text{First term}}$

r = $\frac{8}{1}=8$

Since explicit formula of a geometric sequence is given by

$T_{n}=a(r)^{n-1}$

where $T_{n}$ = nth term of the sequence

a = first term of the sequence

r = common ratio of the successive term to the previous term

Now we plug values of a and r in the formula to get the explicit formula for the given sequence.

$T_{n}=1.(8)^{n-1}$

Therefore, if Bernardo is saying that the formula of the sequence is

h(n) = $1.(8)^{n-1}$ then he is correct.

7 0

2 years ago

Read 2 more answers

If ABCD is a square and AB = 10, what is the measure of CE to the nearest hundredth?

andrey2020 [161]

Answer:

$AC=14.14\ units$

Step-by-step explanation:

<em><u>The correct question is </u></em>

If ABCD is a square and AB = 10, what is the measure of AC? (rounded to the nearest hundredth)

see the attached figure to better understand the problem

we know that

All four sides of the square are congruent and all four interior angles are equal to 90°.

so

AB=BC=CD=AD

In the right triangle ACD

Applying the Pythagorean Theorem

$AC^2=CD^2+AD^2$

we have that

$CD=AD=AB=10\ units$

substitute the given value

$AC^2=10^2+10^2$

$AC^2=200$

$AC=\sqrt{200}\ units$

$AC=14.14\ units$

6 0

2 years ago

Question and answer choices are in the screenshot. THANK YOU!!

Anika [276]

I 11 x - 22 I > 22
∴ 11x-22 > 22          OR          (11x -22) < -22
∴ 11 x > 44             ::::::::          11x - 22 < -22
    x > 4                   ::::::::              x < 0

∴ x ∈ ( -∞ , 0 ) ∪ ( 4 , ∞)
∴ The correct choice is number 1
The solution is attached in the figure

5 0

2 years ago

In a family there are five sons. The oldest son is 11 3/4 years old, and every next younger brother is younger than the next by

timama [110]

Answer:

the youngest brother is of 2 years and 5 months

if this rule still applies, the next son will be born in 2 months.

Step-by-step explanation:

Given that:

Oldest son= 11 3/4 years old = 11 years 3/4 (12) months = 11 years 9 months

Difference = 2 years 7/12(12) months = 2 years 7 moths

So for finding the age of youngest brother:

Subtract the difference others from age of oldest:

= 11 years 9 months - 4(2 years 7 months)

= 11 years 9 months - 9 years 4 months

= 2 years 5 months

So the youngest brother is of 2 years and 5 months

if this rule still applies, the next son will be born in 2 months.

i hope it will help you!

5 0

2 years ago

Read 2 more answers

Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The

Harlamova29_29 [7]

Answer:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

We have the following data given:

$n_1 =20$ represent the sample size for group 1

$n_2 =20$ represent the sample size for group 2

$\bar X_1 =43.5$ represent the sample mean for the group 1

$\bar X_2 =40.1$ represent the sample mean for the group 2

$s_1=4.1$ represent the sample standard deviation for group 1

$s_2=3.2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525$

And the deviation would be just the square root of the variance:

$S_p=3.678$

The statistic is givne by:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0

2 years ago