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Lady bird [3.3K]
3 years ago
7

Which statement about the values 0.034 and 3.40 are true

Mathematics
1 answer:
Nana76 [90]3 years ago
3 0

Answer:

See Explanation

Step-by-step explanation:

<em>This question requires options.</em>

<em>Since none is provided, I will give a general solution.</em>

<em></em>

We have: 0.034 and 3.40

Required

The relationship between them

3.40 can be expressed as:

3.40 \to 0.034 * 100

This means that 3.40 is 100 of 0.034

Similarly, 0.034 can be expressed as:

0.034 \to 3.40 * \frac{1}{100}

This means that 0.034 is 1/100 of 3.40

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At a bake sale, as student spent $11.00 buying 3 brownies and 5 cookies. His friend spent $3.95 buying 1 brownie and 2 cookies.
riadik2000 [5.3K]

Answer:

$$2.25

Step-by-step explanation:

Let b = the cost of one brownie, c = cost of one cookie

3b + 5c = 11

b+2c = 3.95

Times 3 on both sides on the second equation above to get the following:

3b + 6c = 11.85

3b+6c - (3b+5c) = 11.85-11

3b+6c-3b-5c = 0.85

c = 0.85

To find the value of b, use substitution:

b+2c = 3.95

b = 3.95-2c

= 3.95 - 2\times0.85

= 2.25

\therefore the cost of a brownie is $$2.25

Hope this helps :)

4 0
3 years ago
Brainliest if you are correct
frez [133]

Answer:

D) log_2(x)+3log_2(9)

Step-by-step explanation:

Given: log_2(9x^{3})

Use Exponent Rule: 3log_2(9x)

Use Addition Rule: log_2(x)+3log_2(9)

4 0
3 years ago
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What is the value of 1/3x-3/4 when x =1/4
satela [25.4K]

Answer:

The value of 1/3x-3/4 when x=1/4 is 0.08333 repeated.

Step-by-step explanation:

7 0
3 years ago
he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s
sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

P(X

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) =[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

NORMSDIST(6.366)-NORMSDIST(-1.47)

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0
3 years ago
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