All categories

3 years ago

Five years ago Janeta age was one_fifth of her mother'sage .Now she is one_third of her mother's age.Find their present ages?

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Answer:

J - Janeta's age M - mother's age

A) (J-5) = .2 * (M-5)

B) J = (1/3) * M

Subtracting 5 from both sides of B:

B) J -5 = (1/3)M -5

then we substitute the right side of this into the left side of A and we get

(1/3)M -5 = .2 M -1

.333333333333M -5 = .2M -1

.1111111111111111111M = 4

Mother's Age = 36

Janeta's Age = 12

Step-by-step explanation:

You might be interested in

I need help with this urgently!!

galben [10]

Answer:

Step-by-step explanation:

your welcome

3 0

3 years ago

ALGEBRAIC PROOF!!!! PLS HELP!!! The area of a rectangular blanket is 20 square feet. The blanket's length is (2x + 1) ft and the

Fynjy0 [20]

Answer:

Length: 5 ft; width: 4 ft.

Step-by-step explanation:

A = LW formula for area of rectangle

(2x + 1)(2x) = 20 substitute length, width, and area into formula

4x² + 2x - 20 = 0 use the distributive property to multiply out left side

2x² + x - 10 = 0 divide both sides of equation by 2

(2x + 5)(x - 2) = 0 factor out trinomial

2x + 5 = 0 or x - 2 = 0 use zero product rule to solve for x

2x = -5 or x = 2 subtract 5 from both sides; add 2 to both sides

x = -5/2 or x = 2

We discard x = -5/2 since it would give negative length and width, and the length and width cannot be negative.

Length: 2x + 1 = 2(2) + 1 = 5

Width: 2x = 2(2) = 4

Length: 5 ft; width: 4 ft.

7 0

3 years ago

On a standardized exam, the scores are normally distributed with a mean of 165 and astandard deviation of 40. Find the z-score o

vladimir1956 [14]

The Z-score is calculated by the formula below

$\begin{gathered} z-score=\frac{(x-\mu)}{\sigma}_{} \\ \mu=\operatorname{mean} \\ \sigma=s\tan dard\text{ deviation} \\ x=\text{score} \end{gathered}$

Step 2: Substitute the given parameters in the formula

$\begin{gathered} z-\text{score}=\frac{145-165}{40} \\ Z=-\frac{20}{40} \\ Z=-\frac{1}{2} \\ Z=-0.5 \end{gathered}$

Hence, the z-score of a person who scored 145 on the exam is -0.5

8 0

1 year ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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 $s = \frac{\sigma}{\sqrt{n}}$ .