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3 years ago

7

Imani spent half of her weekly allowance playing mini-golf. To earn more money her parents let her wash the car for $4. What is

her weekly allowance if she ended with$12?

1 answer:

Pepsi [2]3 years ago

5 0

Answer:

Well all you have to do is 4x=12 but muiltuply whatever numbers add to 12.such as 3x4=12

Step-by-step explanation:

You might be interested in

What is the area of rectangle with length 18cm and width 4cm?

miv72 [106K]

I hope this helps you

Area=width×length

Area=18×4

Area=72

5 0

3 years ago

Select the equations where x = 8 is the solution. Select all that apply (select 3).

seraphim [82]

Answer:

Step-by-step explanation:

1. x - 2 = 10

x = 10 + 2 = 12

2. 40 = 5x

40/5 = x

x = 8

3. x/2 = 8

Multiply by 2 on both sides

x = 16

4. x + 14 = 24

x = 24 - 14 = 10

5. x + 2 = 10

x = 10 - 2 = 8

6. x/8 = 1

Multiply by 8 on both sides

x = 8

No 2 5 and 6 is the correct answer

3 0

3 years ago

Find the value for x and y for the triangle. Show work.

makkiz [27]

Step-by-step explanation:

done

.......................

7 0

3 years ago

What is the length of AB? Round your answer to the nearest

const2013 [10]

23’3 because I went over the test and I got it correct I’m pretty sure (sorry if I’m wrong I just went with the internet :()

8 0

2 years ago

If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 477 viruses would d

melisa1 [442]

Answer:

0.01596

Step-by-step explanation:

A scientist claims that 8% of the viruses are airborne

Given that:

The population proportion p = 8%

The sample size = 477

We can calculate the standard deviation of the population proportion by using the formula:

$\sigma_p = \sqrt{\dfrac{p(1-p)}{n}}$

$\sigma_p = \sqrt{\dfrac{0.8(1-0.8)}{477}}$

$\sigma_p = \sqrt{\dfrac{0.0736}{477}}$

$\sigma_p = 0.02098$

The required probability can be calculated as:

$P(| \hat p - p| > 0.03) = P(\hat p - p< -0.03 \ or \ \hat p - p > 0.03)$

$= P \bigg ( \dfrac{\hat p -p }{\sqrt{\dfrac{p(1-p)}{n}}} < -\dfrac{0.03}{0.0124} \bigg ) + P \bigg ( \dfrac{\hat p -p }{\sqrt{\dfrac{p(1-p)}{n}}} >\dfrac{0.03}{0.0124} \bigg )$

= P(Z < -2.41) + P(Z > 2.41)

= P(Z < -2.41) + P(Z < -2.41)

= 2P( Z< - 2.41)

From the Z-tables;

$P(| \hat p - p| > 0.03)$ = 2 ( 0.00798

$P(| \hat p - p| > 0.03)$ = 0.01596

Thus, the required probability = 0.01596

3 0

3 years ago