Find the product. (a-6)^2=

5 sweets cost 30p altogether, how much do 7 sweets cost?

Nezavi [6.7K]

The best way to find out what 7 sweets cost, firstly, you should find out what one sweet would cost. You have 5 sweets, with a total cost of 30p. You should divide 30 by 5, giving you 6p per sweet. To get the cost of 7 sweets, you have to multiply 6p by 7, giving you 42p. Therefore, the cost of 7 sweets is 42p. Hope this helps :)

4 0

3 years ago

Read 2 more answers

How many 3/4 ounce spoonfuls of sugar<br> are in a 5 1/2 ounce bowl?<br> show work please

klasskru [66]

Answer: $\frac{22}{3}$

Step-by-step explanation: Divide

$5\frac{1}{2}\div \frac{3}{4}$

Step 1: Convert the mixed number to improper fraction

$5 \frac{1}{2} =\frac{11}{2}$

$= \frac{11}{2}\div \frac{3}{4}$

Step 2: Reciprocal

$\frac{11}{2}\times \frac{4}{3}$

Step 3: Divide the common factors

$4 \div 2 = 2\\\\=\frac{11}{1}\times \frac{2}{3}$

Step 4: Multiply

$\frac{11\times \:2}{1\times \:3}\\\\=\frac{22}{3}$

5 0

3 years ago

How do I solve for x? <br><br>2x + 16 = -5(x + 1)

Vanyuwa [196]

P.E.M/D.A/S PEMDAS multiply subtract 2x from each side them its 16 = -7x -5 then add 5 to each side 21= -7x then divide x = -3

4 0

3 years ago

A study of the amount of time it takes a mechanic to rebuild the transmission for a 1992 Chevrolet Cavalier shows that the mean

Dovator [93]

Answer:

B) 0.0069

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .