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2 years ago

Express 60 as a product of its prime factors

Mathematics

1 answer:

Semenov [28]2 years ago

6 0

Answer:

$2^2\bullet3\bullet5$

Step-by-step explanation:

Write 60 as a product of primes:

↓

$2^2\bullet3\bullet5$

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A rectangular classroom has an area of 300 square feet. The length of the classroom is 5 feet longer than the width, w. Create a

weqwewe [10]

Answer:

w^2 + 5w = 300

Step-by-step explanation:

A= length x width

A =300

Width = w

length = w + 5

Therefore

300 = (w) x (w + 5)

300 = w^2 + 5w

w^2 + 5w = 300

6 0

3 years ago

A mother is now nine times as old as her son. In four years' time she will be five times as old as her son.What are their presen

ivanzaharov [21]

Answer:

son = 4

mom=36

m = 9s

m + 4 = 5(s+4)

m + 4 = 5s + 20

9s + 4 = 5s + 20

4s = 16

s = 4

m = 36

Step-by-step explanation:

5 0

3 years ago

This question please I don't understand this problem either

just olya [345]

Hello,

1)
I suppose we must write correcty and corrects the mistakes!

f(x)=x+3

y=x+3
->x=y+3 exchange x and y
->y=x-3 solve for y

f^(-1)(x)=x-3=g(x)

Here are the mistakes:

(fog)(x)=f(g(x))=f(x-3)=(x-3)+3=x and not -x !

(gof)(x)=g(f(x))=g(x+3)=(x+3)-3=x and not -x!!

f(x) and g(x) ARE inverses each another.

2)
f(x)=1/(4x) x≠0

y=1/(4x)
->x=1/(4y)
->y=1/(4x)
g(x)=f^(-1)(x)=1/(4x)=f(x), x≠0

$(gof)(x)=(fog)(x)=f(g(x))=f(\dfrac{1}{4x})= \dfrac{1}{4* \dfrac{1}{4x}} = \dfrac{1}{ \frac{1}{x}} =1*x=x

$

7 0

4 years ago

Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of bra

Stella [2.4K]

Answer:

The 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =114.1$ represent the sample mean 1

$\bar X_2 =129.9$ represent the sample mean 2

n1=5 represent the sample 1 size

n2=2 represent the sample 2 size

$s_1 =5.08$ sample standard deviation for sample 1

$s_2 =5.37$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =114.1-129.9=-15.8$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_1 +n_2 -1=5+5-2=8$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=\pm 2.31$