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3 years ago

Assistance again, please?

Mathematics

2 answers:

topjm [15]3 years ago

8 0

You're going to need about an 8 feet tall 19 feet wide wrapping paper.

alexira [117]3 years ago

7 0

We're finding how much surface he needs in order to entirely wrap the box, so let's use the surface area form to find it.

$SA: A_1 + A_2 + A_3$

$A_1 = 2(5 \cdot 3)$ (front and back)
$A_2 = 2(3 \cdot 14)$ (two sides)
$A_3 = 2(5 \cdot 14)$ (top and bottom)

Hence, our SA becomes $2(5 \cdot 3) + 2(3 \cdot 14) + 2(5 \cdot 14)$
$= 254 ft^{2}$ of wrapping paper.

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The cost of having photos taken includes a sitting fee and $12 for each portrait. The cost of 3,6, and 9 photos respectively is

Eddi Din [679]

Answer:The sitting fee is 15 dollars

Step-by-step explanation:If you were to use the formula y=mx+b then the starting equation would be y=12x+b. It already gave you the cost for the amount of photos respectively so 51=12(3)+b. 12 times 3 equals 36 and 51 minus 36 equals 15. The same way would be used for 87 dollars and 123 dollars while using the amount of photos respectively. 87=12(6)+b. 12 times 6 equals 72. 72 plus 15 equals 87. 123=12(9)+b. 12 times 9 equals 108. 108 plus 15 equals 123. Therefore the sitting fee is 15. (51=12(3)+15, 87=12(6)+15, 123=12(9)+15)

4 0

3 years ago

If a seed is planted, it has a 95% chance of growing into a healthy plant. If 7 seeds are planted, what is the probability that

denpristay [2]

Answer:

$\frac{51998079}{1280000000}$ or around 0.0406

Step-by-step explanation:

We need to choose 2 seeds that will not grow - there's

7 * 6 / 2 = 21 possibilities (choose one, chose second one, divide by 2 because each choice was counted twice).

The probability of each of those possibilities happening is:

$(5\%)^2 * (95\%)^5 = \frac{2476099}{1280000000}$

because two need to fail to grow and the rest need to grow properly.

The full probability is

$\frac{51998079}{1280000000}$

7 0

3 years ago

Laura borrowed $175,000 at a rate of 5% simple interest for 15 years.How much interest will Laura pay?

never [62]

Answer:

$437.3179611

Step-by-step explanation:

Using the compound interest formula

A = P(1 + $\frac{r}{n}$ ) ^nt

substitute numbers

A = 175,000(1 + $\frac{.05}{15}$ )^ (.05)(15)

solve

7 0

3 years ago

An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6

SIZIF [17.4K]

Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6 times.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 6 trials

r = number of success = at least 4

p = probability of success which in our question is probability of

rolling a “six", i.e; p = 0.20

Let X = Number of sixes on a die

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= $\binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}$

= $15 \times 0.20^{4} \times 0.80^{2}+6 \times 0.20^{5} \times 0.80^{1}+1 \times 0.20^{6} \times 0.80^{0}$

= 0.0154 + 0.00154 + 0.000064

= 0.01696

Therefore, probability of rolling at least 4 sixes is 0.01696.

8 0

3 years ago

a golden eagle flies a distance of 290 miles in 5 days.if the eagle flies the same distance each day of its journey, how far doe

egoroff_w [7]

1450 per day jgjhfjvbjgvnbfhnfg <------ nothing ;)

6 0

3 years ago

Read 2 more answers