Ask question

Ask question

All categories

3 years ago

8

The difference of two numbers is 5. If the sum of four times the first number and the second number is 50, find the numbers. Put

in an ordered pair.

1 answer:

salantis [7]3 years ago

5 0

Answer:

11,6

Step-by-step explanation:

Idk if this is correct

You might be interested in

In a party, there are 10 pairs arriving, each consisting of a boy and a girl. Then, there is a dance, where each boy dances with

tekilochka [14]

Answer:

Expected no. of squads = 184755

Step-by-step explanation:

In a party, there are 10 pairs arriving, each consisting of a boy and a girl.

So that means there are 10 boys and 10 girls.

A squad is defined as a collection of boy-girl pairs and one single pair is also a squad on its own.

We are asked to find out the the expected number of squads.

For 10 pairs of boys and girls:

Number of squads = ₁₀C₁₀ × ₁₀C₁₀ = 1 × 1 = 1

For 9 pairs of boys and girls:

Number of squads = ₁₀C₉ × ₁₀C₉ = 10 × 10 = 100

For 8 pairs of boys and girls:

Number of squads = ₁₀C₈ × ₁₀C₈ = 45 × 45 = 2025

For 7 pairs of boys and girls:

Number of squads = ₁₀C₇ × ₁₀C₇ = 120 × 120 = 14400

For 6 pairs of boys and girls:

Number of squads = ₁₀C₆ × ₁₀C₆ = 210 × 210 = 44100

For 5 pairs of boys and girls:

Number of squads = ₁₀C₅ × ₁₀C₅ = 252 × 252 = 63504

For 4 pairs of boys and girls:

Number of squads = ₁₀C₄ × ₁₀C₄ = 210 × 210 = 44100

For 3 pairs of boys and girls:

Number of squads = ₁₀C₃ × ₁₀C₃ = 120 × 120 = 14400

For 2 pairs of boys and girls:

Number of squads = ₁₀C₂ × ₁₀C₂ = 45 × 45 = 2025

For 1 pairs of boys and girls:

Number of squads = ₁₀C₁ × ₁₀C₁ = 10 × 10 = 100

The expected number of squads is

Expected no. of squads = 1 + 100 + 2025 + 14400 + 44100 + 63504 + 44100 + 14400 + 2025 + 100

Expected no. of squads = 184755

7 0

2 years ago

Can anybody help me with this

ivolga24 [154]

A = (4, 5) B = (-2, 1)

<u>Midpoint of A and B</u>

$C=(X_M, Y_M) = \bigg(\dfrac{X_A+X_B}{2}, \dfrac{Y_A+Y_B}{2}\bigg)\\\\. \qquad \qquad \qquad =\bigg(\dfrac{4-2}{2},\dfrac{5+1}{2}\bigg)\\\\. \qquad \qquad \qquad =\bigg(\dfrac{2}{2},\dfrac{6}{2}\bigg)\\\\. \qquad \qquad \qquad =(1, 3)$

<u>Distance from A to B</u>

$d_{AB}=\sqrt{(X_B-X_A)^2+(Y_B-Y_A)^2}\\\\.\qquad =\sqrt{(-2-4)^2+(1-5)^2}\\\\.\qquad =\sqrt{(-6)^2+(-4)^2}\\\\.\qquad =\sqrt{36+16}\\\\.\qquad =\sqrt{52}\\\\.\qquad =7.2$

<u>Equation of line through AB</u>

$m_{AB}=\dfrac{Y_B-Y_A}{X_B-X_A}\\\\.\qquad =\dfrac{1-5}{-2-4}\\\\.\qquad =\dfrac{-4}{-6}\\\\.\qquad =\dfrac{2}{3}$

$Y-Y_A=m_{AB}(X-X_A)\\\\Y-5=\dfrac{2}{3}(X-4)\\\\Y-5=\dfrac{2}{3}X-\dfrac{8}{3}\\\\Y=\dfrac{2}{3}X+\dfrac{7}{3}$

<u>Line parallel to AB (same slope as AB) through point (3, -5)</u>

$Y-Y_A=m_{AB}(X-X_A)\\\\Y+5=\dfrac{2}{3}(X-3)\\\\Y+5=\dfrac{2}{3}X-2\\\\Y=\dfrac{2}{3}X-7$

AB is <u>perpendicular</u> to A'B' so slopes are <u>opposite reciprocals</u>

A' = (3, 0)

B' = (-1, 6)

C = (1, 3)

C' = (7, 3)

D = (5, 3)

D' = (5, 1)

3 0

2 years ago

In a previous year, 58% of females aged 15 and older lived alone. A sociologist tests whether this percentage is different tod

Gemiola [76]

Answer:

$z=\frac{0.565 -0.58}{\sqrt{\frac{0.58(1-0.58)}{600}}}=-0.744$

Since is a bilateral test the p value would be given by:

$p_v =2*P(z$

And since the p value is higher than the significance level we have enough evidence to conclude that the true proportion is not significantly different from 0.58

Step-by-step explanation:

Information given

n=600 represent the random sample selcted

X=339 represent the number of females aged 15 and older that living alone

$\hat p=\frac{339}{600}=0.565$ estimated proportion of females aged 15 and older that living alone

$p_o=0.58$ is the value that we want to check

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_$ represent the p value

Sytem of hypothesis

We want to check if the true proportion females aged 15 and older that living alone is significantly different from 0.58.:

Null hypothesis: $p=0.58$

Alternative hypothesis: $p \neq 0.58$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.565 -0.58}{\sqrt{\frac{0.58(1-0.58)}{600}}}=-0.744$

Since is a bilateral test the p value would be given by:

$p_v =2*P(z$

And since the p value is higher than the significance level we have enough evidence to conclude that the true proportion is not significantly different from 0.58

5 0

3 years ago

Use the figure shown.

mr Goodwill [35]

Answer:

Step-by-step explanation:

5 0

2 years ago

1. The temperature was -8°F and rose by 2°F. What is the new temperature? Expression:

aev [14]

New temperature: -6°
-8 +2= -6

3 0

2 years ago