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3 years ago

15

Jessica and her 2 sisters want to take a camping trip. They have $225 saved so far. Each of them will save $21 a week until they

have $512 to pay for the trip. How much money will they save after 4 weeks?

1 answer:

Bogdan [553]3 years ago

6 0

Answer:

225 + 63n for n- a natural number,

set 512=225+63n to find total number of weeks

63=21*3

use n=4 to solve the amount saved in 4 weeks

they wont 60*4=240, 240+12, 252+225=477

how much are the saving a day

what do they need to put up to have enough in 4 weeks

how much a day do the need to save to go in 1 week

etc

Step-by-step explanation:

You might be interested in

Find all zeros of x^4+x

Yuri [45]

Answer:

-1 and 0

Step-by-step explanation:

X ^4+x = 0;

x*(x^3+1)=0;

x*[(x^3-x)+(x+1)]=0;

x*[x(x-1)(x+1)+(x+1)]=0;

x*(x+1)(x^2-x+1)=0;

Since x^2-x+1=0 has no intersection with the X-axis, there is no x value such that y=0

x=0 or -1

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2 years ago

the dimensions of kaly's new cell phone are 6 inches by 8 inches. her phone cost 30 per square inch how much did kayla pay for t

Ostrovityanka [42]

Answer:

1440

Step-by-step explanation:

6 x 8 = 48

48 x 30 = 1440

7 0

2 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u> $p_1-p_2$ <u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

<u>95% confidence interval for</u> ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago

A sports stadium has 10000 seats, divided into box seats, lower-deck seats, and upper-devk seats. Box seats sell for 10 dollars,

lesantik [10]

Answer:

Number of box seats = 1000

Number of lower deck seats = 5000

Number of upper deck seats = 4000

Step-by-step explanation:

Let number of box seats = $x$

Let number of lower deck seats = $y$

Let number of upper deck seats = $z$

As per question statement:

$x+y+z=10000$ ...... (1)

Cost for box seat = $10

Cost for $x$ box seats = 10 $x$

Cost for lower deck seat = $8

Cost for $y$ lower deck seats = 8 $y$

Cost for lower deck seat = $5

Cost for $z$ lower deck seats = 5 $z$

As per question statement:

$10x+8y+5z=70000$ ..... (2)

$z = 4x$ ..... (3)

Putting value of $y$ in (1) and (2):

$x+y+4x=10000\\\Rightarrow 5x+y=10000$ ....... (4)

$10x+8 y+5\times 4x=70000\\\Rightarrow 30x+8y=70000 ..... (5)$

8 $\times$ (4) - (5):

$10x =$ 10000

$\Rightarrow x = 1000$

By (3):

$z = 4000$

Now, by equation (1):

$y$ = 5000

Number of box seats = 1000

Number of lower deck seats = 5000

Number of upper deck seats = 4000

6 0

3 years ago

Six people are standing in line for concert tickets. Their average age is 27. The average age of the first four people in line i

scZoUnD [109]

<span>Let the age of the 4th person be F
Sum of the 1st four in line: 4(23), or 92
Sum of 1st 3: 92 - F (excludes the 4th person's age)
Sum of the last 3 in line: 3(34), or 102
</span><span>Sum of last 2: 102 - F (excludes the 4th person's age)
Sum of all 6: 6(27), or 162 (INCLUDES 4th person's age)
We now get: F + 92 - F + 102- F = 162
- F + 194 = 162
- F = 162 - 194
- F = - 32
F, or 4th person's age = -32/-1, or 32

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

</span>

6 0

3 years ago