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3 years ago

OeowowowkwiHELP ASaPieisowowiiwiwiwiiwiwisnsnsnxxj

Mathematics

1 answer:

puteri [66]3 years ago

6 0

Answer:

Step-by-step explanation:

40÷[20-4*(7-4)]

Start with the inner most parentheses

40÷[20-4*(3)]

Then the brackets, multiply first

40÷[20-12]

Then subtract

40÷[8]

We are now left with the division

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What is the present value of $4,900 per year at a discount rate of 10 percent if the first payment is received 6 years from now

Nesterboy [21]

Answer:

17640

Step-by-step explanation:

at a discount of 10% per year $4900 is equal to $490 per year

payment started 6 years ago from now and ended 20 years from now

therefore there is an accumulation of 26 years

for 26 years = 26 *$490= $12740

present value therefore = $4900+$12740=$ 17640

4 0

2 years ago

1. the number 0 (for the real numbers) distributive property 2. a(b + c) = ab + ac or a(b - c) = ab - ac literal equation 3. an

Semenov [28]

the number 0 — additive identity
a(b+c) = ab+ac — the distributive property
an equation containing more than one variable — literal equation
the number 1 — multiplicative identity
the reciprocal of a number — multiplicative inverse

If x ≠ 0, then 1/x is its multiplicative inverse — true.

The product of a number and its multiplicative inverse is 1 — true.

4 0

2 years ago

Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random

Katyanochek1 [597]

Answer:

a) Null hypothesis: $\mu_A =\mu_B =\mu C$

Alternative hypothesis: $\mu_i \neq \mu_j, i,j=A,B,C$

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667$

And we have this property

$SST=SS_{between}+SS_{within}$

The degrees of freedom for the numerator on this case is given by $df_{num}=df_{within}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=3*6-3=15$ .

And the total degrees of freedom would be $df=N-1=3*6 -1 =15$

The mean squares between groups are given by:

$MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166$

And the mean squares within are:

$MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544$

And the F statistic is given by:

$F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326$

And the p value is given by:

$p_v= P(F_{2,15} >11.326) = 0.00105$

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

b) $(\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}$

The degrees of freedom are given by:

$df = n_B +n_C -2= 6+6-2=10$

The confidence level is 99% so then $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ and the critical value would be: $t_{\alpha/2}=3.169$

The confidence interval would be given by:

$(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321$

$(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345$

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part a

Null hypothesis: $\mu_A =\mu_B =\mu C$

Alternative hypothesis: $\mu_i \neq \mu_j, i,j=A,B,C$

If we assume that we have $3$ groups and on each group from $j=1,\dots,6$ we have $6$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667$

And we have this property

$SST=SS_{between}+SS_{within}$

The degrees of freedom for the numerator on this case is given by $df_{num}=df_{within}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=3*6-3=15$ .

And the total degrees of freedom would be $df=N-1=3*6 -1 =15$

The mean squares between groups are given by:

$MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166$

And the mean squares within are:

$MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544$

And the F statistic is given by:

$F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326$

And the p value is given by:

$p_v= P(F_{2,15} >11.326) = 0.00105$

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

Part b

For this case the confidence interval for the difference woud be given by:

$(\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}$

The degrees of freedom are given by:

$df = n_B +n_C -2= 6+6-2=10$

The confidence level is 99% so then $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ and the critical value would be: $t_{\alpha/2}=3.169$

The confidence interval would be given by:

$(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321$

$(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345$

7 0

2 years ago

A spinner has 10 equal-sized sections. To win the game, the pointer must land on a blue section

GrogVix [38]

Answer:

1/10

Step-by-step explanation:

Given data

Total number of possible outcomes /sample space S= 10

In this problem we are required to solve for the probability of obtaining the favourable outcome which is obtaining a blue

Hence the probability of obtaining a blue is

Pr(blue) = 1/10

8 0

2 years ago

Seven and thirty twp hundreths plus two and one tenth times two and seven tenths minus 18a

alexdok [17]

Answer: Write the expression as:

" 70 $\frac{32}{100}$ +
2 $\frac{1}{10}$ * 2 $\frac{7}{10$ − 18a " ;
______________________________________________________
or; write as:
______________________________________________________
" 70.32 + (2.1) * (2.7) − 18a " ;
______________________________________________________

To simplify:
______________________________________________________

Using "PEDMAS" (the "order of operations") ;

the "multiplication" comes first;

So: → "(2.1) * (2.7) = 5.67 " .

And rewrite:
______________________________________________________
" 70.32 + 5.67 − 18a " .

Now: " 70.32 + 5.67 = 75.99 " ;

So, we can the final simplified expression as:
______________________________________________________
" 75.99 − 18a " ;

or; write as: " 75 $\frac{99}{100}$ − 18a " .
______________________________________________________

7 0

3 years ago