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3 years ago

A car moved at a speed of 90km/hr for 3 minutes.What distance did it cover?

Mathematics

1 answer:

aniked [119]3 years ago

5 0

Answer:

270kilometers

Step-by-step explanation:

To find the distance, all you have to do is the speed · time

In these types of distance time and word problems, you multiply the other factors, 90 and 3, The time(T) · The speed(S)

So. 90(S) · 3(T) = 270(D)

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In ΔPQR, the measure of ∠R=90°, the measure of ∠P=26°, and PQ = 8.5 feet. Find the length of QR to the nearest tenth of a foot.

Vitek1552 [10]

Answer:

3.7feet

Step-by-step explanations

Using the sin rule

A/sin a = B/sin b

Let A = PQ = 8.5feet

B = QR = x feet

a = R = 90°

b = P = 26°

Substitute the values into the Sin rule

8.5/sin90 = x/sin26

8.5×sin 26 = x × sin 90

8.5×0.4383 = x× 1

3.7255 = x

Hence the length of QR to the nearest tenth 3.7feet

5 0

2 years ago

Read 2 more answers

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .