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3 years ago

Question 5 of 10

Mathematics

2 answers:

serious [3.7K]3 years ago

7 0

yep. the answer is

$169,000

True [87]3 years ago

4 0

Answer:

The median value of the houses is $169,000

Step-by-step explanation:

Since the houses were undervalued by $15,000, add $15,000 to all amounts

154,000 + 15,000 = 169,000

133,000 + 15,000 = 148,000

167,000 + 15,000 = 182,000

145,000 + 15,000 = 160,000

180,000 + 15,000 = 195,000

172,000 + 15,000 = 187,000

149,000 + 15,000 = 164,000

To take the median of the data set, order the numbers from least to greatest

148,000 160,000 164,000 169,000 182,000 187,000 195,000

Eliminate one number from both sides until you reach one number

The number is 169,000

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You receive an order for 15units of regular insulin to be given with breakfast and 20units to be giving with dinner. Insulin is

Wewaii [24]

Answer:

The medical doctor prescribed 0.35 ml of insulin per day.

Step-by-step explanation:

To find out how many millimiters of insulin was prescribed we need to find out how many units we have to took. The doctor prescribed two does of 15 units and 20 units, so the total of units for the day is the sum of the two in this cas 15 + 20 = 35 units. Now we can use a proportion rule, if we have 100 units for 1 ml in 35 units we will have an x amount of ml:

x = 35/100 = 0.35 ml

The medical doctor prescribed 0.35 ml of insulin per day.

7 0

3 years ago

Read 2 more answers

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

andriy [413]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

(c) Standard deviation of defective light bulbs = 3.67

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

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

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319