Can someone help me with this? I’ll mark you brainliest

The population of a type of local dragonfly can be found using an infinite geometric series where a1 = 65 and the common ratio i

hram777 [196]

In mathematics, number sequencing of the same pattern are called progression. There are three types of progression: arithmetic, harmonic and geometric. The pattern in arithmetic is called common difference, while the pattern in geometric is called common ratio. Harmonic progression is just the reciprocal of the arithmetic sequence.

The common ratio is denoted as r. For values of r<1, the sum of the infinite series is equal to

S∞ = A₁/(1-r), where A1 is the first term of the sequence. Substituting A₁=65 and r=1/6:

S∞ = A₁/(1-r) = 65/(1-1/6)
S∞ = 78

6 0

3 years ago

You lose 5 points for every wrong answer in a trivia game. What integer represents the change in your points after answering 8 q

nata0808 [166]

Your answer would be -40. :-)

4 0

4 years ago

The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated

cluponka [151]

Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

Let $p_1$ = proportion of cockroaches that died on glass surface.

$p_2$ = proportion of cockroaches that died on plasterboard surface.

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of cockroaches that died on glass surface = $\frac{18}{36}$ = 0.50

$\hat p_2$ = sample proportion of cockroaches that died on plasterboard surface = $\frac{25}{36}$ = 0.694

$n_1$ = sample of cockroaches on glass surface = 36

$n_2$ = sample of cockroaches on plasterboard surface = 36

So, test statistics = $\frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36} } }$

= -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

8 0

4 years ago

How many hundreths equal 6 tenths

Tresset [83]

6 tenths = .6 => .60
Then 6 tenth equals to 60 hundredths.

6 0

4 years ago

Write down two factors of 18

777dan777 [17]

Answer:

2 and 9

Step-by-step explanation:

6 0

3 years ago