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3 years ago

girtue goes to the store and buys a rusy car for $55. the store has a sales tax of 5.5% how much will gertue have to pay

Mathematics

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

58.03

Step-by-step explanation:

What I did first to make it easier was make 5.5 to .055 and when you multiply those two together you get 3.025 which if you round it up becomes 3.03 and when you add 55 you get 58.03

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If GHJ is equal to LMK, with a scale factor of 5:6, find the perimeter of GHJ.

xxTIMURxx [149]

Answer: Weeeell, 35

Step-by-step explanation:

1. By definition, the perimeter of a triangle is the sum of the lenght of each side.

2. Then, the perimeter of the triangle LMK is:

P_{LMK}=14+17+11\\P_{LMK}=42

3. If both triangles are congruent and the scale factor is 5:6 (Which you can rewrite as a fraction: 5/6), you must multiply the perimeter of the triangle LMK by this scale factor.

4. Then, you have that the perimeter of the triangle GHJ is:

P_{GHJ}=42*\frac{5}{6}=35

What I have to say:

I hope I helped you, and pls mark brainliest. Have a great day BYYYE!!!!!

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3 years ago

Which intervals are most appropriate for a histogram

Lorico [155]

Answer:

C. 10–15, 15–20, 20–25, 25–30, 30–35

Step-by-step explanation:

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3 years ago

Please help

son4ous [18]

Answer:

Step-by-step explanation:

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3 years ago

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I need help please can someone help me find each area thank you

nikitadnepr [17]

Answer:

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2 years ago

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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago