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1 year ago

Which of the following statements does NOT adequately describe a developmental progression?

1 answer:

6 0

The statement which does not adequately describe a developmental progression is Choice C- A developmental progression is a skill check- list that four-years- olds must master to be prepared for kindergarten.

<h3>What is developmental progression?</h3>

The Developmental Progression of Functional Skills is a concept which describes the child's growth in which case, the child's develops relationship with family members and a sense of self in the process and this concept in discuss is not limited to four-year olds.

Read more on developmental progression;

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Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)

Tems11 [23]

The slope of the tangent line to $f$ at $x=-1$ is given by the derivative of $f$ at that point:

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}$

Factorize the numerator:

$2x^2-2=2(x^2-1)=2(x-1)(x+1)$

We have $x$ approaching -1; in particular, this means $x\neq-1$ , so that

$\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)$

Then

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4$

and the tangent line's equation is

$y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2$

6 0

3 years ago

Can anybody solve this problem?

vaieri [72.5K]

The answer is b! Hope this helped

4 0

3 years ago

Jake is saving for a new tablet. It costs $250, and he has $100. He earns $5 per week for doing chores. Jake wants to figure out

kirza4 [7]

Answer: 250 ≤ 100+5x

150≤5x divide by 5

30≤x

It will take 30 weeks to earn enough.

Step-by-step explanation:

7 0

2 years ago

Cable Strength: A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the ca

KatRina [158]

Answer:

95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].

Step-by-step explanation:

We are given that the engineers take a random sample of 45 cables and apply weights to each of them until they break. The mean breaking weight for the 45 cables is 768.2 lb. The standard deviation of the breaking weight for the sample is 15.1 lb.

Since, in the question it is not specified that how much confidence interval has be constructed; so we assume to be constructing of 95% confidence interval.

Firstly, the Pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean breaking weight = 768.2 lb

s = sample standard deviation = 15.1 lb

n = sample of cables = 45

$\mu$ = population mean breaking strength

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.02 < $t_4_4$ < 2.02) = 0.95 {As the critical value of t at 44 degree

of freedom are -2.02 & 2.02 with P = 2.5%}

P(-2.02 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.02) = 0.95

P( $-2.02 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.02 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.02 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.02 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-2.02 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.02 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $768.2-2.02 \times {\frac{15.1}{\sqrt{45} } }$ , $768.2+2.02 \times {\frac{15.1}{\sqrt{45} } }$ ]

= [763.65 lb , 772.75 lb]

Therefore, 95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].

3 0

3 years ago

Please help me with this problem please please i really need help

Sveta_85 [38]

She paid $35 and part b is it would've been $32

4 0

3 years ago

Read 2 more answers