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3 years ago

15

Ben has typed 180 and can type 30 words per minute. Millie has typed 60 and can type 50 words per minute. How many minutes will

it take for millie to catch up with ben.

1 answer:

Alex17521 [72]3 years ago

4 0

about 2 minutes

hope this helps

You might be interested in

I now have another question to ask.

marissa [1.9K]

Answer:

23.6 ft

Step-by-step explanation:

Sketch a right triangle representing this situation. The length of the hypotenuse is 26 ft and the angle of elevation from ground to top of ladder is 65°. The "opposite side" is the reach of the ladder, which we'll call x.

Then:

opp

sin 65° = ----------
26 ft

or (26 ft)(sin 65°) = opp side = height off the ground of top of ladder.

Evaluating this, we get:

(26 ft)(0.906) = 23.56 ft, or, rounded off, 23.6 ft

The ladder reaches 23.6 ft up the side of the building.

8 0

3 years ago

Which equation involves a prime quadratic and cannot be solved by factoring? A. x2 + 5x − 4 = 0 B. x2 − x − 6 = 0 C. x2 + 3x − 4

Damm [24]

Answer:

Option A

Step-by-step explanation:

We can try to factorize all the options one by one.

For A:

$x^{2} +5x-4\\=> x^{2}+4x+x-4\\=>x(x+4)+1(x-4)$

We can see that the quadratic expression cannot be solved by factorization as the factors at the end of factorization are not equal in both brackets. So Option A is the correct answer for the given question.

Moreover we can also note that all the other quadratic expressions can be factorized ..

8 0

3 years ago

Factor each expression 8+16

astraxan [27]

<span>The individual factors for each integer along with the common factors and greatest common factor will be shown. Factors of 18 are 1, 2, 3, 6, 9, 18. Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.</span>

6 0

2 years ago

Read 2 more answers

The average cost of an IRS Form 1040 tax filing at Thetis Tax Service is $138.00. Assuming a normal distribution, if 77 percent

jek_recluse [69]

Answer:

22.97

Step-by-step explanation:

Given that the average cost of an IRS Form 1040 tax filing at Thetis Tax Service is $138.00.

Let x be the average cost an IRS Form 1040 tax filing at Thetis Tax Service

$P(X$ is given

From std normal distribution table we find 77th percentile z value.

z=0.74

Corresponding X value = 155

i.e. $0.74=\frac{155-138}{s}$ where s is the std deviation

Simplify to get

Std deviation = $\frac{17}{0.74} \\=22.97$

8 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