All categories

4 years ago

Devaughn is 15 years younger than Sydney. The sum of their ages is 73. What is Sydney’s age?

Mathematics

2 answers:

Olegator [25]4 years ago

6 0

Answer:

Sydney is 44 years old

Step-by-step explanation:

Step 1: Make a system of equations

d = s - 15

d + s = 73

Step 2: Plug in s - 15 for d in the second equation and solve

s - 15 + s = 73

2s - 15 = 73 + 15

2s / 2= 88 / 2

s = 44

Answer: Sydney is 44 years old

Alex73 [517]4 years ago

3 0

Answer:

Sydney is 44. I think, hope this helps!

You might be interested in

A company rounds its losses to the nearest dollar. The error on each loss is independently and uniformly distributed on [–0.5, 0

lesya [120]

Answer:

the 95th percentile for the sum of the rounding errors is 21.236

Step-by-step explanation:

Let consider X to be the rounding errors

Then; $X \sim U (a,b)$

where;

a = -0.5 and b = 0.5

Also;

Since The error on each loss is independently and uniformly distributed

Then;

$\sum X _1 \sim N ( n \mu , n \sigma^2)$

where;

n = 2000

Mean $\mu = \dfrac{a+b}{2}$

$\mu = \dfrac{-0.5+0.5}{2}$

$\mu =0$

$\sigma^2 = \dfrac{(b-a)^2}{12}$

$\sigma^2 = \dfrac{(0.5-(-0.5))^2}{12}$

$\sigma^2 = \dfrac{(0.5+0.5)^2}{12}$

$\sigma^2 = \dfrac{(1.0)^2}{12}$

$\sigma^2 = \dfrac{1}{12}$

Recall:

$\sum X _1 \sim N ( n \mu , n \sigma^2)$

$n\mu = 2000 \times 0 = 0$

$n \sigma^2 = 2000 \times \dfrac{1}{12} = \dfrac{2000}{12}$

For 95th percentile or below

$P(\overline X < 95}) = P(\dfrac{\overline X - \mu }{\sqrt{{n \sigma^2}}}< \dfrac{P_{95}- 0 } {\sqrt{\dfrac{2000}{12}}}) =0.95$

$P(Z< \dfrac{P_{95} } {\sqrt{\dfrac{2000}{12}}}) = 0.95$

$P(Z< \dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}}) = 0.95$

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1- 0.95$

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} = 0.05$

From Normal table; Z > 1.645 = 0.05

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1.645$

${P_{95}\sqrt{12} } = 1.645 \times {\sqrt{{2000}}}$

${P_{95} = \dfrac{1.645 \times {\sqrt{{2000}}} }{\sqrt{12} } }$

$\mathbf{P_{95} = 21.236}$

the 95th percentile for the sum of the rounding errors is 21.236

8 0

3 years ago

A study was conducted to compare the proportion of drivers in Boston and New York who wore seat belts while driving. Data were c

allochka39001 [22]

Answer:

a) The test statistic is $z = -2.51$

b) The pvalue is 0.0060.

Step-by-step explanation:

Question a:

Ha: pB < pNY

This means that the alternate hypothesis is rewritten as:

$H_a: p_{B} - p_{NY} < 0$

While the null hypothesis is:

$H_0: p_{B} - p_{NY} = 0$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis and s is the standard error

pB = 0.581, pNY = 0.832

This means that $X = 0.581 - 0.832 = -0.251$

0 is tested at the null hypothesis:

This means that $\mu = 0$

Standard error = 0.1

This means that $s = 0.1$

Test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{-0.251 - 0}{0.1}$

$z = -2.51$

The test statistic is $z = -2.51$

Question b:

The pvalue is the probability of finding a proportion less than -0.251, that is, a difference of at least 0.251, which is the pvalue of Z = -2.51.

Looking at the z-table, Z = -2.51 has a pvalue of 0.0060

The pvalue is 0.0060.

5 0

3 years ago

Identify the trinomial that is a perfect square.

nignag [31]

Answer:

the perfect square trinomial formula is: a 2 ± a b + b 2 a^{2} \pm ab + b^{2} a2±ab+b2.

8 0

3 years ago

Prove Sin3theta = 3sintheta - 4sin^3theta

trapecia [35]

Express the left hand side as

sin3theta=sin(theta+2theta)

now expand the right side of this equation using color(blue)"Addition formula"

color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))

rArrsin(theta+2theta)=sinthetacos2theta+costhetasin2theta.......(A)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta)color(white)(a/a)|)))

The right hand side is expressed only in terms of sintheta's

so we use cos2theta=1-2sin^2theta........(1)

color(red)(|bar(ul(color(white)(a/a)color(black)(sin2theta=2sinthetacostheta)color(white)(a/a)|)))........(2)

Replace cos2theta" and " sin2theta by the expansions (1) and (2)
into (A)

sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)

and expanding brackets gives.

sin(theta+2theta)=sintheta-2sin^3theta+2sinthetacos^2theta....(B)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1rArrcos^2theta=1-sin^2theta)color(white)(a/a)|)))

Replace cos^2theta=1-sin^2theta" into (B)"

rArrsin(theta+2theta)=sintheta-2sin^3theta+2sintheta(1-sin^2theta)

and expanding 2nd bracket gives.

sin(theta+2sintheta)=sintheta-2sin^3theta+2sintheta-2sin^3theta

Finally, collecting like terms.

sin3theta=3sintheta-4sin^3theta="R.H.S hence proven"

3 0

3 years ago

A total of 487 tickets were sold for the school play. They were either adult tickets or student tickets. There were 63 fewer stu

shepuryov [24]

S=a-63

s+a=487, using s from above in this equation you get:

a-63+a=487 combining like terms

2a-63=487 adding 63 to both sides

2a=550 dividing both sides by 2

a=275

So there were 275 adult tickets sold.

7 0

3 years ago