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Alekssandra [29.7K]

3 years ago

14

Which division problem would have an estimated quotient of 15 if the divisor and dividend were rounded to the nearest whole numb

er? 90.24 divided by 6.3 79.85 divided by 3.99 85.01 divided by 16.77 59.87 divided by 10.21

2 answers:

jarptica [38.1K]3 years ago

4 0

Answer:

Option a) 90.24 divided by 6.3

Step-by-step explanation:

We take each option one by one and solve for it

a) 90.24 divided by 6.3

Divisor = 90.24

Dividend = 6.3

if the divisor and dividend were rounded to the nearest whole number?

= 90÷ 6

= 15

This option is correct.

b) 79.85 divided by 3.99

79.85 = Divisor

3.99= Dividend

if the divisor and dividend were rounded to the nearest whole number?

= 80 ÷ 4

= 20

This is wrong

c) 85.01 divided by 16.77

85.01 = Divisor

16.77 = Dividend

if the divisor and dividend were rounded to the nearest whole number?

= 85÷17

= 5

This option is wrong

d) 59.87 divided by 10.21

Divisor = 59.87

Dividend = 10.21

if the divisor and dividend were rounded to the nearest whole number?

= 60÷ 10

=6

This option is wrong.

Therefore, the division problem that would have an estimated quotient of 15 if the divisor and dividend were rounded to the nearest whole number is Option a) 90.24 divided by 6.3

melamori03 [73]3 years ago

4 0

Answer: Hey yall wassup it's me Tj and thats period um chile anyways let me get straight to it. the answer really be A) 6.3/90.24

Step-by-step explanation: Yall stay pretty and classy and HAWTTTT

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A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou

Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let $z$ be the

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question, $x = 5$ and $\sigma = 2$ . The equation becomes

$\displaystyle z = \frac{5 - \mu}{2}$ .

To solve for $\mu$ , the mean of this distribution, the only thing that needs to be found is the value of $z$ . Since

The problem stated that $P(X \le 5) = 69.85\% = 0.6985$ . Hence, $P(Z \le z) = 0.6985$ .

The problem is that the normal distribution tables list only the value of $P(0 \le Z \le z)$ for $z \ge 0$ . To estimate $z$ from $P(Z \le z) = 0.6985$ , it would be necessary to find the appropriate

Since $P(Z \le z) = 0.6985$ and is greater than $P(Z \le 0) = 0.50$ , $z > 0$ . As a result, $P(Z \le z)$ can be written as the sum of $P(Z < 0)$ and $P(0 \le Z \le z)$ . Besides, $P(Z < 0) = P(Z \le 0) = 0.50$ . As a result:

$\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}$ .

Therefore:

$\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}$ .

Lookup $0.1985$ on a normal distribution table. The corresponding $z$ -score is $0.52$ . (In other words, $P(0 \le Z \le 0.52) = 0.1985$ .)

Given that

$z = 0.52$ ,
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To figure out how much weight the pig gained, we must find 30% of 120.

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Therefore, the pig gained 36 pounds.

To find the pig’s new weight, we must add 36 pounds to the pig’s original 120 pound weight.

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Therefore, the new weight of Shane’s pig is 156 pounds.

Hope this helps!

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