Ask question

Ask question

All categories

2 years ago

15

If the Townsends have a monthly budget of $3000 and spend 20% of that on food,how much is spent on food??????? I am in desparate

need of help!!!! Got a big test today!! How do I find out how to find it??? Please help!!!

1 answer:

andre [41]2 years ago

7 0

3000 x .20 = 600 would be the answer

You might be interested in

A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha

ollegr [7]

Answer:

99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

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

where, $\hat p$ = sample proportion of students received a passing grade = 77%

n = sample of tests = 40

p = population proportion

<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u>99% confidence interval for p</u> = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ , $0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ ]

= [0.5986 , 0.9414]

Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Lower bound of interval = 0.5986

Upper bound of interval = 0.9414

6 0

2 years ago

Solve the equation identify extraneous solutions

PIT_PIT [208]

Answer:

Step-by-step explanation:

The goal to solving any equation is to have x = {something}. That means we need to get the x out from underneath that radical. It's a square root, so we can "undo" it by squaring. Square both sides because this is an equation. Squaring both sides gives you

$x^2=-3x+40$

Get everything on one side of the equals sign and set the quadratic equal to 0:

$x^2+3x-40=0$

Throw this into the quadratic formula to get that the solutions are x = 5 and -8. We need to see if only one works, both work, or neither work in the original equation.

Does $5=\sqrt{-3(5)+40}$ ?

$5=\sqrt{-15+40}$ and

$5=\sqrt{25}$

and 5 = 5. So 5 works. Let's try -8 now:

$-8=\sqrt{-3(-8)+40}$ and

$-8=\sqrt{24+40}$ so

$-8=\sqrt{64}$

-8 = 8? No it doesn't. So only 5 works. Your choice is the third one down.

6 0

3 years ago

What is the Value of f(6) if f(x)=2x-11

il63 [147K]

The answers are
f
o
g
(
x
)
=
−
2
x
+
23
and
g
o
f
(
x
)
=
−
2
x
+
5

Explanation:
f
(
x
)
=
−
2
x
+
11

g
(
x
)
=
x
−
6

f
o
g
(
x
)
=
f
(
g
(
x
)
)
=
f
(
x
−
6
)
=
−
2
(
x
−
6
)
+
11

=
−
2
x
+
12
+
11
=
−
2
x
+
23

g
o
f
(
x
)
=
g
(
f
(
x
)
)
=
g
(
−
2
x
+
11
)
=
−
2
x
+
11
−
6

=
−
2
x
+
5

I think that the equations speak by themselves.
Of course,
f
o
g
(
x
)
≠
g
o
f
(
x
)

3 0

2 years ago

Read 2 more answers

The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and

Firdavs [7]

Answer:

1a

$P(39 < X < 48 ) = 0.8767$

1b

95% of all sample means will fall between $40.1 < \mu < 49.9$

1c

$\= x = 41. 795$

2

$n = 25$

Step-by-step explanation:

From the question we are told that

The mean is $n = 45$

The population standard deviation is $\sigma = 10$

The sample size is n = 16

Generally the standard error of the mean is mathematically represented as

$\sigma_{x} = \frac{ \sigma}{\sqrt{n} }$

=> $\sigma_{x} = \frac{ 10 }{\sqrt{16 } }$

=> $\sigma_{x} = 2.5$

Generally the probability that the sample mean will be between 39 and 48 minutes is

$P(39 < X < 48 ) = P( \frac{ 39 - 45}{ 2.5} < \frac{X - \mu }{\sigma } < \frac{ 48 - 45}{ 2.5} )$

=> $P(39 < X < 48 ) = P(-2.4 < Z< 1.2 )$

=> $P(39 < X < 48 ) = P( Z< 1.2 ) - P(Z < -2.4)$

From the z table the area under the normal curve to the left corresponding to 1.2 and -2.4 is

=> $P( Z< 1.2 ) = 0.88493$

and

$P( Z< - 2.4 ) = 0.0081975$

So

$P(39 < X < 48 ) = 0.88493 -0.0081975$

=> $P(39 < X < 48 ) = 0.8767$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * 2.5$

=> $E =4.9$

Generally the 95% of all sample means will fall between

$\mu -E < and \mu +E$

=> $45 -4.9\ and \ 45 + 4.9$

Generally the value which 90% of sample means is greater than is mathematically represented

$P( \= X > \= x ) = 0.90$

=> $P( \= X > \= x ) = P( \frac{\= X - \mu }{ \sigma_x} > \frac{\= x -45 }{ 2.5} ) = 0.90$

=> $P( \= X > \= x ) = P( Z > z ) = 0.90$

Generally from the z-table the critical value of 0.90 is

$z = -1.282$

$\frac{\= x -45 }{ 2.5} = -1.282$

=> $\= x = 41. 795$

Considering question 2

Generally we are told that the standard deviation of the mean to be one fifth of the population standard deviation, this is mathematically represented as

$s = \frac{1}{5} \sigma$

Generally the standard deviation of the sample mean is mathematically represented as

$s = \frac{\sigma }{ \sqrt{n} }$

=> $\frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }$

=> $n = 5^2$

=> $n = 25$

5 0

2 years ago

3. Damien paints 1/5 of a room in 5/12 hours what unit rate describes this situation

sdas [7]

Answer:

It takes 2 1/2 hours to paint each room.

7 0

2 years ago