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3 years ago

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A graphing calculator is recommended. For the limit lim x → 3 (x3 − 4x + 3) = 18 illustrate the definition by finding the larges

t possible values of δ that correspond to ε = 0.2 and ε = 0.1. (Round your answers to four decimal places.) ε = 0.2 δ = ε = 0.1 δ =

1 answer:

alexira [117]3 years ago

6 0

Answer:

Attached is the detailed solution of the problem

∈ = 0.0081 ≈ 0.2

∈ = 0.0042 ≈ 0.1

Step-by-step explanation:

Attached is the detailed solution of the problem

∈ = 0.0081 ≈ 0.2

∈ = 0.0042 ≈ 0.1

A graphing calculator recommended was used to arrive at this solution

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At a charity function, $150,045 was donated by three companies A, B, and C in the ratio 7 : 5 : 9. How much money did company B

Svetach [21]

35,725
this is because u can take the ratio 7+5+9=21 then take 150,045/21 which u get 7,145 then time 5 and u get 35,725

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3 years ago

Read 2 more answers

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58

Olegator [25]

Answer:

Hypothesis Test states that we will accept null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, $\mu_1$ = true mean voltage for type K batteries.

$\mu_2$ = true mean voltage for type Q batteries.

So, Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, $H_1$ : $\mu_1 \neq \mu_2$ {mean voltage for these two types of

batteries is different]

<em>The test statistics we use here will be :</em>

$\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ follows $t_n__1 + n_2 -2$

where, $X_1bar$ = 8.54 and $X_2bar$ = 8.69

$s_1$ = 0.225 and $s_2$ = 0.725

$n_1$ = 37 and $n_2$ = 58

$s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2} }{37+58-2} }$ = 0.585 Here, we use t test statistics because we know nothing about population standard deviations.

Test statistics = $\frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58} } }$ follows $t_9_3$

= -1.219

<em>At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.</em>

Therefore, we conclude that mean voltage for these two types of batteries is same.

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3 years ago

Find three consecutive natural numbers if the square of the smallest number is 65 less than the product of the two remaining num

garri49 [273]

Answer:

21 22 23

Step-by-step explanation:

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3 years ago

Help me I need it asap<br>

mel-nik [20]

Answer:53-6

Step-by-step explanation:

That’s the answer

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3 years ago

Many states assess the skills of their students in various grades. One program that is available for this purpose is the Nationa

Nataly [62]

Answer:

A score of 314 is needed to be in the top 25% of students who take this exam.

Step-by-step explanation:

We are given that one of the tests provided by the NAEP assesses the reading skills of twelfth-grade students. In recent years, the national mean score was 289 and the standard deviation was 37.

Let X = <u><em>scores of the tests provided by the NAEP</em></u>

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = national mean score = 289

$\sigma$ = standard deviation = 37

Now, we have to find how high a score is needed to be in the top 25% of students who take this exam, that means;

P(X > x) = 0.25 {where x is the required score}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-289}{37}$ ) = 0.25

P(Z > $\frac{x-289}{37}$ ) = 0.25

In the z table, the critial value of z that represents the top 25% of the area is given as 0.6745, that is;

$\frac{x-289}{37}=0.6745$

${x-289}=0.6745\times 37$

x = 289 + 24.96 = 313.96 ≈ 314

Hence, a score of 314 is needed to be in the top 25% of students who take this exam.

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3 years ago