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Nitella [24]
2 years ago
15

Francis makes a trip to the mall to purchase clothes. He has a $100

Mathematics
2 answers:
faust18 [17]2 years ago
4 0

Answer:

<em>$9.98 is how much more money is needed to make the purchases</em>.

Step-by-step explanation:

First, you would add the cost of the shirt ($29.99) and the cost of the shoes ($79.99),

$29.99 + $79.99 = $109.98

then...... subtract the budget ($100.00) from the total cost of the shirt and shoes ($109.98),

$109.98 - $100.00 = $9.98

$9.98 is the amount of money still needed to make the purchases.

Alex2 years ago
3 0

Answer:

$9.98

Step-by-step explanation:

Amount of more money required to make the purchase:-

=>(cost of shirt + cost of pair of shoes) - budget

=>(79.99+29.99)-100

=>109.98-100

=><u>9.98</u>

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The solution to 5−3x &gt; 35 is either x &gt; -10 or -10 &gt; x. Which solution is correct? Explain how you know.
STatiana [176]

Answer:

x > -10

Step-by-step explanation:

5-3x > 35

-3x > 30

x > -10

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How do you turn a wood that is 8 feet that has been cut into 6 pieces into a fraction
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The board is 8/6 which is also 1 1/3
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2 years ago
Triangle ABC was dilated using the rule Y, 5/4. FCA is equal to eight what is C’A’ 10 units 12 and 16 units 20 units
seropon [69]

Answer:

C'A' = 10units (A)

Question

A complete question related to this found at brainly(question ID 2475535) is stated below.

Triangle ABC was dilated using the rule Dy, 5/4

If CA = 8, what is C'A'?

10 units

12 units

16 units

20 units

Step-by-step explanation:

Given:

Scale factor = 5/4

CA = 8units

Find attached the diagram for the question.

This is a question on dilation. In dilation, figures have the same shapes but different sizes.

Y is the center of dilation

Lengths of ∆ABC: CB, AB, CA

Lengths of ∆A'B'C': C'B', A'B', C'A'

C'B' = scale factor × CB

A'B' = scale factor × AB

C'A' = scale factor × CA

C'A' = 5/4 × 8

C'A' = 40/4

C'A' = 10units (A)

8 0
2 years ago
The stiffness of a certain type of steel beam used in building construction has mean 30 kN/mm and standard deviation 2 kN/mm. Wh
Alecsey [184]

Answer:

15.87% probability that the average of 100 randomly selected beams is greater than 30.2 kN/mm

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 30, \sigma = 2, n = 100, s = \frac{2}{\sqrt{100}} = 0.2

What is the probability that the average of 100 randomly selected beams is greater than 30.2 kN/mm?

This probability is 1 subtracted by the pvalue of Z when X = 30.2. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{30.2 - 30}{0.2}

Z = 1

Z = 1 has a pvalue of 0.8413.

1 - 0.8413 = 0.1587

15.87% probability that the average of 100 randomly selected beams is greater than 30.2 kN/mm

5 0
2 years ago
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