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3 years ago

Which equation can be used to find the two numbers whose ratio is 3 to 2 and that have a sum of 35

Mathematics

2 answers:

Gnoma [55]3 years ago

7 0

Three times two equals six x

3x2=6x

dusya [7]3 years ago

4 0

Idk like 3+2 _=5
35
--- = 7
5
3:2
21:14

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Drag the tiles to the correct boxes to complete the pairs.

Ne4ueva [31]

Answer:

Part 1) $-1.25$ -------> $2.75/(-2.2)$

Part 2) $-4\frac{1}{3}$ --------> $(-2\frac{3}{5}) / (\frac{3}{5})$

Part 3) $\frac{2}{3}$ ------> $(-\frac{10}{17}) / (-\frac{15}{17})$

Part 4) $3$ ------> $(2\frac{1}{4}) / (\frac{3}{4})$

Step-by-step explanation:

Part 1) we have

$2.75/(-2.2)$

To calculate the division problem convert the decimal number to fraction number

$2.75=275/100\\ -2.2=-22/10$

$(275/100)/(-22/10)$

Remember that

Since division is the opposite of multiplication, you can turn this division problem into a multiplication problem by multiplying the top fraction by the reciprocal of the bottom fraction

$(275/100)/(-22/10)=(275/100)*(-10/22)=-(275*10)/(22*100)=-(275)/(220)$

Simplify

Divide by 22 both numerator and denominator

$-(275)/(220)=-125/100=-1.25$

Part 2) we have

$(-2\frac{3}{5}) / (\frac{3}{5})$

To calculate the division problem convert the mixed number to an improper fraction

$(-2\frac{3}{5})=-\frac{2*5+3}{5}=-\frac{13}{5}$

$(-\frac{13}{5}) / (\frac{3}{5})$

Since division is the opposite of multiplication, you can turn this division problem into a multiplication problem by multiplying the top fraction by the reciprocal of the bottom fraction

$(-\frac{13}{5}) / (\frac{3}{5})=(-\frac{13}{5})*(\frac{5}{3})=-\frac{13*5}{5*3}=-\frac{13}{3}$

Convert to mixed number

$-\frac{13}{3}=-(\frac{12}{3}+\frac{1}{3})=-4\frac{1}{3}$

Part 3) we have

$(-\frac{10}{17}) / (-\frac{15}{17})$

Since division is the opposite of multiplication, you can turn this division problem into a multiplication problem by multiplying the top fraction by the reciprocal of the bottom fraction

$(-\frac{10}{17}) / (-\frac{15}{17})=(-\frac{10}{17})*(-\frac{17}{15})=\frac{10*17}{17*15}=\frac{10}{15}$

Simplify

Divide by 5 both numerator and denominator

$\frac{10}{15}=\frac{2}{3}$

Part 4) we have

$(2\frac{1}{4}) / (\frac{3}{4})$

To calculate the division problem convert the mixed number to an improper fraction

$(2\frac{1}{4})=\frac{2*4+1}{4}=\frac{9}{4}$

$(\frac{9}{4}) / (\frac{3}{4})$

Since division is the opposite of multiplication, you can turn this division problem into a multiplication problem by multiplying the top fraction by the reciprocal of the bottom fraction

$(\frac{9}{4}) / (\frac{3}{4})=(\frac{9}{4})*(\frac{4}{3})=\frac{9*4}{4*3}=\frac{9}{3}=3$

8 0

3 years ago

On a single roll of a pair of dice, what are the odds against rolling a sum of 7? Please help ASAP!!! :(

Kobotan [32]

Honestly I don’t remember how to do this but u can try .083 but don’t count on it

7 0

2 years ago

Minimum and maximum value of 2n^2+5n-25=0

Alborosie

Equations don't have minimum or maximum, functions do.

Function y=2n^2+5n-25 has minimum -28.125, has no maximum.

8 0

3 years ago

Whats the figures and how do I find the area?

iragen [17]

Answer:

This is trapezium.

Area of this figure =1/2h(d1-d2)

1/2×5(14-8)

=30/2

=15cm^2

8 0

3 years ago

An article in Fire Technology investigated two different foam-expanding agents that can be used in the nozzles of firefighting s

UNO [17]

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to test if two different foam-expanding agents have the same foam expansion capacity

Sample 1 (aqueous film forming foam)

n₁= 5

X[bar]₁= 4.7

S₁= 0.6

Sample 2 (alcohol-type concentrates )

n₂= 5

X[bar]₂= 6.8

S₂= 0.8

Both variables have a normal distribution and σ₁²= σ₂²= σ²= ?

The statistic to use to make the estimation and the hypothesis test is the t-statistic for independent samples.:

t= $\frac{(X[bar]_1 - X[bar]_2) - (mu_1 - mu_2)}{Sa*\sqrt{\frac{1}{n_1} + \frac{1}{n_2 } } }$

a) 95% CI

(X[bar]_1 - X[bar]_2) ± $t_{n_1 + n_2 - 2}$ * $Sa* \sqrt{\frac{1}{n_1}+\frac{1}{n_2} }$

Sa²= $\frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1 + n_2 - 2}$ = $\frac{(5-1)0.36 + (5-1)0.64}{5 + 5 - 2}$ = 0.5

Sa= 0.707ç

$t_{n_1 + n_2 -2: 1 - \alpha /2} = t_{8; 0.975} = 2.306$

(4.7-6.9) ± 2.306* $(0.707\sqrt{\frac{1}{5}+\frac{1}{5} })$

[-4.78; 0.38]

With a 95% confidence level you expect that the interval [-4.78; 0.38] will contain the population mean of the expansion capacity of both agents.

The hypothesis is:

H₀: μ₁ - μ₂= 0

H₁: μ₁ - μ₂≠ 0

α: 0.05

The interval contains the cero, so the decision is to reject the null hypothesis.

<u>Complete question</u>

a. Find a 95% confidence interval on the difference in mean foam expansion of these two agents.

b. Based on the confidence interval, is there evidence to support the claim that there is no difference in mean foam expansion of these two agents?

8 0

3 years ago