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3 years ago

If one-third of one-fourth of a number is 15, then three-tenth of that number is:

Mathematics

1 answer:

aksik [14]3 years ago

6 0

Answer:

Step-by-step explanation:

Let the number be x.

1/3 of 1/4 of x = 15 <=> 3 x 4 x 15 = x = 180

3/10 x 180 = 54

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Decide whether each table could represent a proportional relationship. If the

Ymorist [56]

Answer:

Step-by-step explanation:

If given tables in the picture show the proportional relationship,

Number of wheels (w) ∝ Number of buses (b)

w ∝ b

w = kb

Here, k = proportionality constant

k = $\frac{w}{b}$

Number of buses (b) Number wheels (w) Wheels per bus $(\frac{w}{b})$

5 30 $\frac{30}{5}=6$

8 48 $\frac{48}{8}=6$

10 60 $\frac{60}{10}=6$

15 90 $\frac{90}{15}=6$

Here, proportionality constant is 6.

Similarly, If number of wheels (w) ∝ Number of train cars (t)

w = kt

Here, k = proportionality constant

k = $\frac{w}{t}$

Number of train cars(t) Number of wheels(w) Wheels per train car ( $\frac{w}{t}$ )

20 184 $\frac{184}{20}=9.2$

30 264 $\frac{264}{30}=8.8$

40 344 $\frac{344}{40}=8.6$

50 424 $\frac{424}{50}=8.48$

Since, ratio of w and t is not constant, relation between number of wheels and number of train cars is not proportional.

5 0

2 years ago

Find the 63rd term of the arithmetic sequence -1, 5, 11, ...−1,5,11,...

Ugo [173]

Answer:

$a_6_3=371$

Step-by-step explanation:

we know that

In an <u><em>Arithmetic Sequence</em></u> the difference between one term and the next is a constant and this constant is called the common difference

we have

$-1,5,11,...$

Let

$a_1=-1\\a_2=5\\a_3=11$

$a_2-a_1=5-(-1)=5+1=6$

$a_3-a_2=11-5=6$

The common difference is $d=6$

We can write an Arithmetic Sequence as a rule

$a_n=a_1+d(n-1)$

where

a_n is the nth term

d is the common difference

a_1 is the first term

n is the number of terms

Find the 63rd term of the arithmetic sequence

we have

$n=63\\d=6\\a_1=-1$

substitute

$a_6_3=-1+6(63-1)$

$a_6_3=-1+6(62)$

$a_6_3=-1+372$

$a_6_3=371$

6 0

3 years ago

By what percentages are Sweden's high category and medium category less than the United States' categories (to the nearest perce

Korolek [52]

REQUIRED CHART :

The required chart has been attached

Answer:

31.8%

30.0%

Step-by-step explanation:

Required :

To obtain the Difference between Sweden and United States high and medium categories to the nearest %

Sweden :

High category = $23.51

Medium category = $15.73

United States :

High category = $34.48

Medium category = $22.46

Percentage Difference :

High category : (34.48 - 23.51) / 34.48 * 100% = 31.8%

Medium category : (22.46 - 15.73) / 22.46 * 100% = 29.96% = 30.0%

7 0

2 years ago

Mr. Johnson buys 5 bottles of lemonade for the school picnic. He buys four 28

Murljashka [212]

Answer:

180

Step-by-step explanation:

112+64

=112+64

=176

4 0

2 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

uysha [10]

Answer:

$(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743$

$(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926$

And we are 9% confidence that the true mean for the difference of the population means is given by:

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

Step-by-step explanation:

For this problem we have the following data given:

$\bar X_1 = 9$ represent the sample mean for one of the departments

$\bar X_2 = 8$ represent the sample mean for the other department

$n_1 = 25$ represent the sample size for the first group

$n_2 = 20$ represent the sample size for the second group

$s_1 = 2$ represent the deviation for the first group

$s_2 =1$ represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

The confidence given is 95% or 9.5, then the significance level is $\alpha=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n_1 +n_2 -2= 20+25-2= 43$

And the critical value for this case is $t_{\alpha/2}=2.02$

And replacing we got:

$(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743$

$(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926$

And we are 9% confidence that the true mean for the difference of the population means is given by:

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

4 0

2 years ago