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tamaranim1 [39]

1 year ago

13

Roger has 4 gallóns of orange juice. He puts the same amount of juice each of 5 pitchers. How many gallons of orange juice are i

n 1 pitcher?

1 answer:

salantis [7]1 year ago

5 0

Answer:

1.25

Step-by-step explanation:

you divide 5 within 4 to get how much Roger puts on each pitch

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A ranger in a lookout tower spots two fires on the campground below. Fire A is 75 meters east

Wittaler [7]

Answer:

the fire hydrant is 9 meters west and 58 m south of the tower

Step-by-step explanation:

Fire A and Fire B can be represented as a point in the Cartesian coordinate with the tower as the origin. Fire A is 75 meters east and 40 meters south of the tower, It can be represented as (75, -40). Fire B is 37 meters west and 64 meters south of the tower, It can be represented as (-37, -64).

If a point O(x, y) divides a line segment AB in the ratio n:m, with A( $x_1,y_1$ ) and B( $x_2,y_2$ ), the point O(x, y) is at:

$x=\frac{n}{n+m}(x_2-x_1)+x_1\\ \\y=\frac{n}{n+m}(y_2-y_1)+y_1$

The fire hydrant (x, y) divides Fire A(75, -40) and fire B(-37, -64) in the ratio 3:1(three fourth), hence:

$x=\frac{n}{n+m}(x_2-x_1)+x_1=\frac{3}{4}(-37-75) +75=\frac{3}{4}(-112)+75=-84+75\\x=-9 \\ \\y=\frac{n}{n+m}(y_2-y_1)+y_1=\frac{3}{4}(-64-(-40))+(-40)= \frac{3}{4}(-18)-40=-18-40\\y=-58$

The fire hydrant is at (-9, -58). That is the fire hydrant is 9 meters west and 58 m south of the tower

8 0

1 year ago

Let P2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by 10x2+4xâ1, 3xâ4x2+3, and

lord [1]

I suppose

$H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}$

The vectors that span $H$ form a basis for $P_2$ if they are (1) linearly independent and (2) any vector in $P_2$ can be expressed as a linear combination of those vectors (i.e. they span $P_2$ ).

Independence:

Compute the Wronskian determinant:

$\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0$

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of $H$ is 3.

Span:

Write an arbitrary vector in $P_2$ as $ax^2+bx+c$ . Then the given vectors span $P_2$ if there is always a choice of scalars $k_1,k_2,k_3$ such that

$k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c$

which is equivalent to the system

$\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}$

so the vectors do span $P_2$ .

The vectors comprising $H$ form a basis for it because they are linearly independent.

4 0

2 years ago

PART A: Marvin has a coupon the discount is the rental of a full-size car by $25 they decide to buy insurance for each day if th

natita [175]

Answer:

i dont know about the other parts but part A is 19

Step-by-step explanation:

6 0

1 year ago

A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr

erastovalidia [21]

Answer:

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

$p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034$

The estimated standard error is:

$\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316$

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

$LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096$

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

8 0

1 year ago

Find the sum of 67.008 and 3.8

zysi [14]

Answer:

69.808

Step-by-step explanation:

69.808

4 0

2 years ago