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velikii [3]
4 years ago
11

Each month Adam changes the amount of food and sunlight a potted plant receives. When he started the plant was 63 cm tall. In th

e first month it grew 14 cm. The following month the plant grew 3 times as much as it did the first month. How many meters tall is the plant at the end of the second month?
Mathematics
2 answers:
iren [92.7K]4 years ago
4 0
At the end of the second month, the plant is 1.19 meters tall.
Umnica [9.8K]4 years ago
4 0
The plant is 119 m tall at the end of the second month.
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JOe goes to the gym to run and swim. When running he burns 35
OverLord2011 [107]

Answer:the length of time he spent running is 8 minutes.

the length of time he spent swimming is 15 minutes.

Step-by-step explanation:

Let x represent the length of time he spent running.

Let y represent the length of time he spent swimming.

He exercised for a total of 23 minutes. This means that

x + y = 23

When running he burns 35 calories per minute, and when he swims he burns 30 calories per minute. He has burned 730 calories after exercising. It means that

35x + 30y = 730 - - - - - - - - - -1

Substituting x = 23 - y into equation 1, it becomes

35(23 - y) + 30y = 730

805 - 35y + 30y = 730

- 35y + 30y = 730 - 805

- 5y = - 75

y = - 75/ -5

y = 15

Substituting y = 15 into x = 23 - y, it becomes

x = 23 - 15 = 8

8 0
3 years ago
AA3 +2=AAA CC6+6=CBB ABC=?
Ahat [919]
AA3 + 2 = AAA; 

<span><span>3 + 2 = 5; </span><span>- >> A = 5; </span><span>AAA = 555. </span></span>


<span>
CC6 + 6 = CBB; </span>
<span><span>6 + 6 = 12; </span><span>- >> B = 2; </span><span>one goes to the left and C (1 +?) turns into B; </span><span>and B already found = 2; </span></span>

<span><span>C = B-1 = 2-1 = 1; </span><span>- >>> C = 1; </span><span>SVB = 122; </span></span>


<span><span>
ABC =? </span><span>- >> 521.

</span></span>
5 0
3 years ago
Ax+z=aw-y solve for a
kodGreya [7K]
I don’t think you wrote the problem correct, I’m a freshmen in college and your problem is stating A*X+Z=A*w-y to solve this problem we would need to know the givens. I may be completely spacing on this being the fact a tool a gap year before going back to school but this doesn’t seem right, sorry
8 0
4 years ago
Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents wh
kherson [118]

Answer:

a. The alternative hypothesis H₀: p'₁ ≠ p'₂ is accepted

b. Type I error

Step-by-step explanation:

Proportion of California residents who reported insufficient rest = 8.0%

Proportion of Oregon  residents who reported insufficient rest = 8.8%

p'₁ = 0.08 * 11545 =923.6

p'₂ = 0.088 * 4691=412.81

σ₁ = \sqrt{n*p_1*q_1}  = \sqrt{n*p_1*(1-p_1)} = \sqrt{11545*0.08*(1-0.08)} = 29.15

σ₂ = \sqrt{n*p_2*q_2}  = \sqrt{n*p_2*(1-p_2)}= \sqrt{4691*0.088*(1-0.088)} = 19.40

Samples size of California residents n₁ = 11,545

Samples size of Oregon residents n₂ = 4,691

Hypothesis can be constructed thus

Let our null hypothesis be H ₀: p'₁ = p'₂

and alternative hypothesis H ₐ: p'₁ ≠ p'₂

Then we have  

z =\frac{(p'_1 -p'_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2} } }

The test statistics can be computed by

     

t₀ = \sqrt{\frac{n_1n_2(n_1+n_2-2)}{n_1+n_2} } *\frac{p_1'-p_2'}{\sqrt{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2} } =      1104.83

c from tables is   P(T ≤ c) = 1 - α where α = 5% and c = 1.65

since t₀ ≥ c then then the hypothesis is rejected which means the alternative hypothesis is rejected

b. Type I error, rejecting a true hypothesis

5 0
3 years ago
Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0
Vsevolod [243]

Answer:

\dfrac{-1}{6}

Step-by-step explanation:

Given the limit of a function expressed as \lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}, to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)

Step 2: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the function

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\

Step 3: substitute x = 0 into the resulting function

= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)

Step 4: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\

=  \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)

Step 6: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\

Step 7: substitute x = 0 into the resulting function in step 6

=  \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}

<em>Hence the limit of the function </em>\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \  is \ \dfrac{-1}{6}.

3 0
3 years ago
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