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4 years ago

Answer ASAP please −0.1·(100a+10b−c)

Mathematics

2 answers:

Aneli [31]4 years ago

8 0

Answer:

- 10a-b+0.1c

Step-by-step explanation:

−0.1·(100a+10b−c)= - 10a-b+0.1c

AleksAgata [21]4 years ago

7 0

Answer:

-10a-b+0.1c

Step-by-step explanation:

−0.1·(100a+10b−c)

-0.1×(100a+10b-c)=

-10a-b+0.1c

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Grades in large classes are sometimes approximately Normally distributed—a fact that serves as the justification for "grading on

QveST [7]

Answer:

Grade A: $Z \geq 1$

Grade B: $0 \leq Z < 1$

Grade C: $-1 \leq Z < 0$

Grade D: $Z < -1$

Step-by-step explanation:

Problems of normally distributed samples can be solved using the Z score table.

The Z score of a measure represents how many standard deviations it is above or below the mean of all the measures.

Each Z score has a pvalue. This represents the percentile of the measure.

In this problem, we have that:

The upper 16% of the class get A grades. The upper 16% has a pvalue of at least 100% = 16% = 84% = 0.84. This is $Z \geq 1$ .

The middle 34% of the class get B grades. The middle 34% has a pvalue of at least 84%-35% = 50% = 0.5 and at most 0.84. This is $0\leqZ < 1$ .

Those between a pvalue of 0.5-0.34 = 0.16 and 0.5 get get grade C. $Z = -1$ has a pvalue of 0.16. So a grade C is in the interval $-1 \leq Z < 0$ .

Those with Z lesser than -1 get grades D and F

6 0

3 years ago

The perimeter of an isosceles triangle is 14. If the

VashaNatasha [74]

Answer:

The length of each leg = 5 units

The length of the base = 4 units

Step-by-step explanation:

Isosceles Triangle is a triangle with any two sides of it in equal measure.

Perimeter of a triangle = SUM OF ALL SIDES

Now, Perimeter of the triangle = 14

Let length of each leg = a

So, the length of the other leg is also a.

And, the base of the triangle is a -1

Now, as perimeter is 14.

⇒a + a + (a -1) = 14

or, 3 a = 14 + 1

⇒ 3 a = 15, or a = 15/3 = 5

Hence, the length of each leg = 5

And the length of the base = (a -1) = (5-1) = 4 units

6 0

4 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05