Please help me if you know the answer. Id like fr you to explain. Not recommended but I would like it.

(t+5)(t+2),make t the subject of the formula

lianna [129]

Answer:

t= 2p+5/p-1

Step-by-step explanation:

The question is not well writen

Assume p = t+5/t-2

We are to make t the subject of the formula

Cross multiply

p(t-2) = t+5

pt - 2p = t+5

pt - t = 2p+5

t(p-1) = 2p+5

t= 2p+5/p-1

Hence the value of t is 2p+5/p-1

5 0

3 years ago

A yeast culture weighing 2 grams is removed from a refrigerator unit and is expected to grow at the rate of Upper W prime (t )eq

Naily [24]

Answer:

a) during the first 10 hours of growth the weight will increase 271.8% relative to the initial state ( or 2.718 g)

b) during the first 20 hours of growth the weight will increase 467.1 % relative to the initial state ( or 4.671 g)

Step-by-step explanation:

since the growing rate law is

W'(t) = 0.1 gr/hour * e^(0.1gr/hour* t) , W'(t) [gr/hour]

and following mathematical conventions: W'(t)= dW/dt

then

dW/dt=0.1 e^(0.1t)

∫dW =∫0.1 e^(0.1t) dt

W = e^(0.1t) + C

at the beginning, (time t=0) the weight is W=2 grams .Therefore

2 g = e^(0.1 g/h*0) + C → 2 g = 1 g + C → C = 1 g

then

W = e^(0.1t) + 1 g

at t= 10 hours

W = e^(0.1 g/h*10h) + 1 g = 3.718 g/h

therefore the weight will increase

ΔW = 3.718 g - 1 g = 2.718 g or 271.8% relative to the initial state

for t=20 hours

W = e^(0.1 g/h*20h) + 1 g = 8.389 g/h

thus, the from t= 10 hours to t= 20 hours the weight will increase

ΔW = 8.389 g/h - 3.718 g = 4.671 g or 467.1 %relative to the initial state

5 0

3 years ago

Dupe age and olu age add up to 25 years eight years ago,Dupe was twice as old as olu.How old are they?

lana66690 [7]

Answer:

dupe is 20, olu is 5

Step-by-step explanation:

3 0

3 years ago

A university is interested in whether there's a difference between students who live on

ryzh [129]

It is to be noted that the determination of whether or not there is a significant difference between the two groups will be done using a t test.

A t-test is a statistical test that juxtaposes two samples' means. It is used in hypothesis testing, using a null hypothesis that the variance in group means is zero and an alternative hypothesis that the difference is not zero.

<h3>What are the conditions for using this kind of confidence interval? </h3>

The conditions to use the t test are:

The sample must be independent
The mean of the population and variance must be unknown.
The Box plot is attached.

<h3>What are the degrees of freedom (k) for this test using the conservative method?</h3>

The degrees of freedom (k) to be utilized for this text will be derived using the conservative method given below:

df = [(s₁²/n₁) + (s₂²/n²)/[((s₁²/n₁)²/((n₁-1)) + (s₂²/n₂)²/((n₂-1))]

= [(3.0952/15) + (6.4095/15)]² / [((3.0952/15)²/14) + ((6.4095/15)²/14)]

= 24.965

Hence,

df ≈ 24 (if approximated to the floor)

<h3>What are the sample statistics for this test?</h3>

Recall the the standard deviation of the population are unequal and unknown. This thus requires that we utilize the two-sample unpooled t-test.

Here, H₀ is given as;

$t = \frac{\bar{x_{1} -\bar{x_{2}}}}{\sqrt{\frac{s_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \sim t_{df}$

t = [(3.33333 - 4.13333)]/√[(3.0952/15) + (6.4095/15)]

= - 0.8/√0.6337

t = - 1.005

<h3>What is the 95% confidence interval for the difference between the number of classes missed by each group of students?</h3>

The 95% confidence interval is computed using the following formula:

$(\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \right)$

= - 0.8 ± t₀.₀₂₅,₂₄ (√0.6337)

= - 0.8 ± 2.064 (√0.6337)

= -2.4429, 0.8429

<h3>What is the a 90% confidence interval for the difference between the number of classes missed by each group of students?</h3>

To derive the 90% interval, we state:

$(\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \right)$

= - 0.8 ± t₀.₀₅₀,₂₄ (√0.6337)

= - 0.8 ± 0.685 (√0.6337)

= -2.162, 0.562

<h3>Based on the two confidence intervals computed in parts d and e, what is the conclusion about the differences between the means of the two groups?</h3>

From the intervals computed, we must fail to reject H₀

H₀ : μ₁ = μ₂

It is clear from the above intervals computed from that the differences between the mean of both groups is significant. This is because, zero is included on the two intervals.

Learn more about t-test at;
brainly.com/question/6589776
#SPJ1

8 0

2 years ago

I need help with this question???

Pepsi [2]

Answer: Zero slope.

Step-by-step explanation:

There is no " Rise over run "

5 0

4 years ago