All categories

3 years ago

The mean of a,b,c,and d is 9 and the mean of a,b,c,d,e and f is 12. find the mean of e and f

Mathematics

1 answer:

Simora [160]3 years ago

8 0

Answer:

mean of e + f = 18

Step-by-step explanation:

(a + b + c + d)/4 = 9 Multiply by 4

a + b + c + d = 9 * 4 = 36

(a + b + c + d + e + f ) / 6 = 12

but a + b + c + d = 36

(36 + e + f)/6 = 12 Multiply by 6

36 + e + f = 72 Subtract 36 from both sides

e + f = 36 Divide by 2 to find the mean.

(e + f)/2 = 36/2

mean of e + f = 18

You might be interested in

Five more than the product of a number and 4 is equal to 8

Maru [420]

Answer:

$4x+5=8$

$x=\frac{3}{4}$

Step-by-step explanation:

Let

x ---> the number

we know that

The algebraic expression is equal to multiply the number by 4, add 5 and equal 8

$4x+5=8$

Solve for x

subtract 5 both sides

$4x=8-5\\4x=3$

Divide by 4 both sides

$x=\frac{3}{4}$

3 0

3 years ago

Solve for x and y. x= y=

lianna [129]

Answer:

x=1

y=7

I mean it's just a guess

6 0

3 years ago

PQRS has vertices P(0, 0), Q(4, 2), and S(4, ␣2). Its diagonals intersect at H(4, 0). What are the possible coordinates of R for

True [87]

Any point on the x-axis beyond the point (4, 0) is a possible cordinate of R.

8 0

3 years ago

A national study report indicated that 20.9% of Americans were identified as having medical bill financial issues. What if a ne

Alenkasestr [34]

Answer:

The p-value for this test is 0.22065.

Step-by-step explanation:

We are given that a national study report indicated that 20.9% of Americans were identified as having medical bill financial issues.

A news organization randomly sampled 400 Americans from 10 cities and found that 90 reported having such difficulty.

Let p = proportion of Americans who were identified as having medical bill financial issues in 10 cities.

SO, Null Hypothesis, $H_0$ : p $\leq$ 20.9% {means that % of Americans who were identified as having medical bill financial issues in these 10 cities is less than or equal to 20.9%}

Alternate Hypothesis, $H_A$ : p > 20.9% {means that % of Americans who were identified as having medical bill financial issues in these 10 cities is more than 20.9% and is more severe}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 400 Americans from 10 cities who were found having such difficulty = $\frac{90}{400}$ = 0.225 or 22.5%

n = sample of Americans = 400

So, test statistics = $\frac{0.225-0.209}{{\sqrt{\frac{0.225(1-0.225)}{400} } } } }$

= 0.77

Now, P-value of the test statistics is given by the following formula;

P-value = P(Z > 0.77) = 1 - P(Z $\leq$ 0.77)

= 1 - 0.77935 = 0.22065

5 0

3 years ago

HELPPPPPPP how many solutions does this have

Fiesta28 [93]

I need the pic or question to solve

7 0

3 years ago

The mean of a,b,c,and d is 9 and the mean of a,b,c,d,e and f is 12. find the mean of e and f​

The mean of a,b,c,and d is 9 and the mean of a,b,c,d,e and f is 12. find the mean of e and f