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4 years ago

How do I solve this math problem?

Mathematics

2 answers:

notsponge [240]4 years ago

5 0

Teresa had more points because since the fractions are not able to use you can see that 6 is larger than 5

Furkat [3]4 years ago

5 0

Teresa's book had more pages than Abigail's. By one page

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Women's shoe size follow a normal distribution with a mean of size 7.5 and a standard deviation of 1.5 sizes. Using the 68-95-99

Alexus [3.1K]

Given:
μ = 7.5, the population mean
σ = 1.5, the population standard deviation

According to the 68 - 95 - 99.7 rule, 68% of the population data lies within one standard deviation of the population mean.
Therefore the range of sizes that fit 68% of women is
(7.5-1.5, 7.5+1.5) = (6.0, 9.0)

Answer: 6.0 to 9.0

7 0

3 years ago

Francois baked just enough cookies to fill all the orders at his bakery. While the cookies were cooling, a kitchen assistant kno

galben [10]

I'm guessing the question is how many total cookies were baked.
36 = .12 x *** x being the total number of cookies
Divide both sides by .12
300 = x

7 0

3 years ago

Read 2 more answers

How can you use the prime factorization of the powers of ten to find the prime factorization of 270,000?

RoseWind [281]

Prime factorization is the process of finding the factors of a number and these factors must be of prime numbers.

27000 = 27 × 10000
27000 = 27 × 10⁴
27000 = 3 × 9 × 10⁴ [Here we have 3 as a prime number, but not '9' and '10', so we will keep breaking this down]
27000 = 3 ×3 × 3 × (2 × 5) ⁴ [Here we have all prime numbers]
27000 = 3³ × 2⁴ × 5⁴

8 0

4 years ago

Find The derivative of:(cosx/1+sinx)^3

forsale [732]

Answer:

$-\frac{3 \cdot cos^2x}{(1+sinx)^3}$

Step-by-step explanation:

$y = (\frac{cosx}{1+sinx})^3\\\\\frac{dy}{dx} = 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{dy}{dx}(\frac{cosx}{1+sinx})$ $[ y = x^n\ \ \ => \ \ \ \frac{dy}{dx} = b \cdot x^{n-1} \ ]$

$= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{(1+sin x(-sinx) - cosx(cosx)}{(1+sinx)^2}\\\\$ $[\ \frac{u}{v} = \frac{v \dcot u'- u \cdot v'}{v^2}\ ]$

$= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x-sin^2x- cos^2x}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x- (sin^2x+ cos^2x)}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x-1}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-1 \cdot(sin x+1)}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-1}{(1+sinx)}\\\\$

$= -3 \cdot \frac{cos^2x}{(1+sinx)^3}$

5 0

3 years ago

A sample of five measurements, randomly selected from a normally distributed population, resulted in the following summary stati

Murrr4er [49]

Answer:

a. null hypothesis is rejected.

b. null hypothesis is rejected.

For (a); 0.05 < P < 0.100

For (b) 0.100 < P < 0.200

Step-by-step explanation:

Given

Number of samples, n = 5

Mean, x = 4.8

Standard Deviation, s = 1.3

I'm finding it difficult to submit my answer

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7 0

3 years ago