(fractions) 8/9 + 3/18

Mathematics

1 answer:

SVEN [57.7K]2 years ago

5 0

Answer:

13/18

Step-by-step explanation:

s t u p i d

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The first term of an arithmetic sequence is 7 and the sixth term is 22. Find(a)the common difference;(2)(b)the twelfth term;(2)(

SVEN [57.7K]

d = 3 , a₁₂ = 40 and S $x_{100}$ = 7775

In an arithmetic sequence the nth term and sum to n terms are

<h3>• a $n$ = a₁ + (n-1)d</h3><h3>• S $_n$ = $\frac{n}{2}$ [2a + (n-1)d]</h3><h3>where d is the common difference</h3><h3>a₆ = a₁ + 5d = 22 ⇒ 7 + 5d = 22 ⇒ 5d = 15 ⇔ d = 3</h3><h3>a₁₂ = 7 + 11d = 7 +( 11× 3) = 7 + 33 = 40</h3><h3>S₁₀₀ = $\frac{50}{2}$ [(2×7) +(99×3)</h3><h3> = 25(14 + 297) = 25(311)= 7775</h3>

7 0

2 years ago

NEED HELP ASAP!! IF YOU ANSWER RIGHT ILL MARK YOU BRAINLIEST!!

goblinko [34]

Y = 5x + 4
Gradient = slope = 5
(0,4) = y - intercept

7 0

2 years ago

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w

dimulka [17.4K]

Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$