Translate the following sentence into an equation. Use x to represent an unknown number. Do not solve.

A shopper pushes a cart 50 m south down one aisle and then turns 90.0° west and moves 23 m. He then makes another 90.0° turn and

irga5000 [103]

Answer:

Step-by-step explanation:

Given

Shopper pushes a cart 50 m south down then turns 90 ^{\circ} west and moves 23 m.

he then makes $90^{\circ}$ turn and moves 14 m

(a)Largest possible displacement is when he turns towards south

Net displacement $=\sqrt{64^2+23^2}=68.0 m$

(b)Smallest displacement when he turns towards north

$=\sqrt{36^2+23^2}$

$=\sqrt{1825}=42.72 m$

6 0

3 years ago

Factor the following trinomial expression by a method of your choice: 3x2 + 7x-10.

polet [3.4K]

3x² + 7x - 10
3x² - 3x + 10x - 10
3x(x) - 3x(1) + 10(x) - 10(1)
3x(x - 1) + 10(x - 1)
(3x + 10)(x - 1)

The answer is D.

5 0

3 years ago

Read 2 more answers

Five players are given $10 each and asked to contribute any portion of it to a group account. They are also told that the total

Natasha2012 [34]

Answer:

nothing

Step-by-step explanation:

4 0

2 years ago

g In a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained

Marat540 [252]

Answer:

95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

Step-by-step explanation:

We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.

Firstly, the Pivotal quantity for 95% 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 residents 65-years or older who had obtained a flu shot = $\frac{28}{111}$ = 0.25

n = sample of residents 65-years or older = 111

p = population proportion of residents who were getting the flu shot

Here for constructing 95% confidence interval we have used One-sample z test for proportions.

So, 95% confidence interval for the population proportion, p is ;

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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ , $0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ ]

= [0.169 , 0.331]

Therefore, 95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

7 0

3 years ago

Who answers this correctly with explanation gets vevmo'd/pay pal 25 Any Links Will Be Reported.

Inga [223]

Answer:

the answer is C

hope this helps

have a good day :)

Step-by-step explanation:

4 0

3 years ago