All categories

3 years ago

PLEASE HEL MUST BE DONE URGENTLY

Mathematics

2 answers:

LekaFEV [45]3 years ago

8 0

Y = x + 2 is closest to what I believe is the solution.

If the club begins with 2 members, then +2 MUST appear in the formula.

If the club adds 2 members every week, then the total number of members would be

f(x) = y = 2 + 2(x-1), where x is the number of weeks.

Test this: If x = 1, then y = 2+2(1-1) = 2 + 0 = 2. This is correct. The club began with 2 members in week 1

marysya [2.9K]3 years ago

8 0

Hi there!

We know that on week 1, there are two students.

That means when x=1, y should be 2.

After testing each answer choice, you can eliminate every single answer choice except for y=2x.

The "2" before the x in this equation means that y is increasing 2 units when x increases 1 unit, which matches the given information.

Thus, y = 2x should be your answer.

Have an awesome day! :)

You might be interested in

Round 0.42 to the nearst tenth

Archy [21]

0.4 because if it’s five or more, add one more. If it’s four or less, let it rest.

4 0

3 years ago

PLS HELP SOLVE FOR X

horsena [70]

Answer:

x=26

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Prove that √2 +√5 is irrational

Sindrei [870]

We have to prove that $\sqrt{2}+\sqrt{5}$ is irrational. We can prove this statement by contradiction.

Let us assume that $\sqrt{2}+\sqrt{5}$ is a rational number. Therefore, we can express:

$a=\sqrt{2}+\sqrt{5}$

Let us represent this equation as:

$a-\sqrt{2}=\sqrt{5}$

Upon squaring both the sides:

$(a-\sqrt{2})^{2}=(\sqrt{5})^{2}\\a^{2}+2-2\sqrt{2}a=5\\a^{2}-2\sqrt{2}a=3\\\sqrt{2}=\frac{a^{2}-3}{2a}$

Since a has been assumed to be rational, therefore, $\frac{a^{2}-3}{2a}$ must as well be rational.

But we know that $\sqrt{2}$ is irrational, therefore, from equation $\sqrt{2}=\frac{a^{2}-3}{2a}$ the expression $\frac{a^{2}-3}{2a}$ must be irrational, which contradicts with our claim.

Therefore, by contradiction, $\sqrt{2}+\sqrt{5}$ is irrational.

4 0

3 years ago

Calcium is an essential nutrient for strong bones and for controlling blood pressure and heart beat. Because most of the body’s

irakobra [83]

Answer:

No, because the 95% confidence interval contains the hypothesized value of zero.

Step-by-step explanation:

Hello!

You have the information regarding two calcium supplements.

X₁: Calcium content of supplement 1

n₁= 12

X[bar]₁= 1000mg

S₁= 23 mg

X₂: Calcium content of supplement 2

n₂= 15

X[bar]₂= 1016mg

S₂= 24mg

It is known that X₁~N(μ₁; σ²₁), X₂~N(μ₂;δ²₂) and σ²₁=δ²₂=?

The claim is that both supplements have the same average calcium content:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

The confidence level and significance level are to be complementary, so if 1 - α: 0.95 then α:0.05

since these are two independent samples from normal populations and the population variances are equal, you have to use a pooled variance t-test to construct the interval:

[(X[bar]₁-X[bar]₂) ± $t_{n_1+n_2-2;1-\alpha /2}$ * $Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ ]

$t_{n_1+n_2-2;1-/2}= t_{25;0.975}= 2.060$

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} }= \sqrt{\frac{11*529+14*576}{12+15-2} } = 23.57$

[(1000-1016)±2.060*23.57* $\sqrt{\frac{1}{12} +\frac{1}{15} }$ ]

[-34.80;2.80] mg

The 95% CI contains the value under the null hypothesis: "zero", so the decision is to not reject the null hypothesis. Then using a 5% significance level you can conclude that there is no difference between the average calcium content of supplements 1 and 2.

I hope it helps!

3 0

3 years ago

Difference quotient

Nina [5.8K]

$f(x)=5x^2-5x-6
\\ \quad \\

\begin{cases}
f(\boxed{x+h})=5(\boxed{x+h})^2-5(\boxed{x+h})-6
\end{cases}\qquad thus
\\ \quad \\
\cfrac{f(x+h)-f(x)}{h}\qquad \textit{will be then}
\\ \quad \\
\cfrac{[5({x+h})^2-5({x+h})-6]\quad -\quad [5x^2-5x-6]}{h}
\\ \quad \\
\cfrac{[5(x^2+2xh+h^2)-5(x+h)-6]-[5x^2-5x-6]}{h}
$

$\cfrac{\underline{5x^2}+10xh+5h^2\underline{-5x}-5h\underline{-6}\underline{-5x^2}\underline{+5x}\underline{+6}}{h}\impliedby \textit{canceling those ones}
\\ \quad \\
\cfrac{10xh+5h^2-5h}{h}\impliedby \textit{common factor}
\\ \quad \\
\cfrac{5\underline{h}(2x+h-1)}{\underline{h}}$

and surely, you'd know what that is

6 0

3 years ago