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nataly862011 [7]
3 years ago
15

Crystal's mud mask costs $23 for a 0.45 oz jar. She found the same mud mask online for $72. What is the size of the jar Crystal

found online?
Mathematics
1 answer:
adell [148]3 years ago
5 0
Because .45 divided by 23 is .0195621739
That is how much it is per dollar
That answer times 72 is 1.408695652
So the size of the jar is about 1.41 oz
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a lighthouse casts a 128 footshadow a nearby lampost that measures 5 feet 3inches cast an 8 Foot Shadow
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5 times 3 times 8 is the correct answer, because it's like finding the volume.<span />
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Find the percent decrease. Round to the nearest percent.<br><br><br> From 90°F to 70°F
Goryan [66]

Answer:

22

Step-by-step explanation:

90*x = 70

x = 7/9

x = 78%

100-78 = 22

3 0
3 years ago
What is the true solution to the logarithmic equation below?<br> log₂ (6x)-log₂ (√x)=2
slega [8]

Answer:

The true solution is x=4/9

EXPLANATION

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When we apply this law the expression becomes

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6 0
2 years ago
Read 2 more answers
A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), <em>p</em> = 0.52

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        =P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

=P(X

Thus, the probability that the flight has empty seats is 0.4625.

4 0
3 years ago
What is the value of x?<br> 6.75+1= 13.<br> &lt;“<br> 7<br> 18<br> O 2<br> O 171<br> O 182<br> O 535
lana66690 [7]

The following answer is 171

4 0
3 years ago
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