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3 years ago

10-24

Mathematics

1 answer:

jasenka [17]3 years ago

8 0

Answer:

Like this...

Step-by-step explanation:

The best represented graph should look something like this :)

(program used is Desmos)

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PLEASE ANSWER ASAP

posledela

This is how you set up a percent of error proportion:
Amount of error x(Unknown percent)
Actual value = 100(Percents are out of hundreds)

So if you fill it in, it becomes 4 x
 80 =100
Cross multiply, 400= 80x.
Solve.
x=5 so percent error is 5%

5 0

3 years ago

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 83 type K batteries and a sample of 77

agasfer [191]

Answer:

1. 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]

2. Test Statistics value = -5.06

4. Decision for the hypothesis test is that we will reject 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]

The test statistics we use here will be :

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

where, $X_1bar$ = 9.29 and $X_2bar$ = 9.65

$s_1$ = 0.374 and $s_2$ = 0.518

$n_1$ = 83 and $n_2$ = 77

$s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(83-1)0.374^{2}+(77-1)0.518^{2} }{83+77-2} }$ = 0.45 Here, we use t test statistics because we know nothing about population standard deviations.

Test statistics = $\frac{(9.29-9.65) - 0 }{0.45\sqrt{\frac{1}{83}+\frac{1}{77} } }$ follows $t_1_5_8$

= -5.06

At 0.05 or 5% level of significance t table gives a critical value between (-1.98,-1.96) to (1.98,1.96) at 158 degree of freedom. Since our test statistics is less than the critical table value of t as -5.06 < (-1.98,-1.96) so we have sufficient evidence to reject null hypothesis.

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

7 0

4 years ago

Is Q the midpoint or PR? please justify the answer

mezya [45]

Since PQ=PR
=> Q is the midpoint of PR
Hope this helps you