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3 years ago

Plssssss help number 18 and 20

Mathematics

2 answers:

weqwewe [10]3 years ago

6 0

Answer:

Total mass of materials = 4.8kg

<u>For</u><u> </u><u>gravel</u>

55% of the materials are gravel

So we have

55/100 × 4.8 kg

= 2.64kg

<u>For</u><u> </u><u>sand </u>

30% of the materials are sand

We have

30/100 × 4.8

= 1.44kg

<u>For</u><u> </u><u>cement</u>

The remaining materials were cement so we add the masses of both the sand and gravel and subtract it from the total mass

That's

4.8 - (2.64 + 1.44)

4.8 - 4.08

= 0.72 kg

She earns £ 800 per week

She receives a raise of 7.5% which is

100 + 7.5 = 107.5%

So we have

107.5 / 100 × £ 800

= £ 860

She now earns £ 860 per week

Hope this helps you

Xelga [282]3 years ago

5 0

Step-by-step explanation:

1)of 4.8kg concrete 55% is gravel,30% is sand and 15% is cement.

percent=Rate*Base

Pgravel=55%*4.8kg

Pgravel=2.64kg

Psand=30%*4.8kg

Psand=1.44kg

Pcement=15%*4.8kg

Pcement=0.72kg

2)percent=Rate*Base

percent=7.5%*£800

percent=60£

total earn=60£ + £800

total earn=£860

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Suppose that insurance companies did a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle

nirvana33 [79]

Answer:

i) $ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$ME= 1.96*\sqrt{\frac{0.8 (1-0.8)}{400}} =0.0392$

ii) $Lower = 0.8-0.0392= 0.7608$

$Upper = 0.8 +0.0392= 0.8392$

Step-by-step explanation:

Notation and definitions

$X=320$ number of people that claimed always buckle up.

$n=400$ random sample taken

$\hat p=\frac{320}{400}=0.8$ estimated proportion of people that claimed always buckle up

$p$ true population proportion of people that claimed always buckle up

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

Part i

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The margin of error is given by:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$ME= 1.96*\sqrt{\frac{0.8 (1-0.8)}{400}} =0.0392$

Part ii

And the confidence interval would be given by:

$Lower = 0.8-0.0392= 0.7608$

$Upper = 0.8 +0.0392= 0.8392$

6 0

3 years ago

Person A can paint the neighbors house 2 times as fast as Person B. The year A and B worked together, it took them 2 days. How l

seraphim [82]

Answer:

A can paint in 3days and B in 6 days

Step-by-step explanation:

From the question we were told

Person A can paint the neighbors house 2 times as fast as Person B

Let A only paint the neighbors house in X days

Let B only paint the neighbors house in 2X days

If we express the given information from the question in terms of fraction we have,

1/X + 1/2X = 1/2

3/2X = 1/2

if we cross multiply we have

6= 2X

X= 6/2

X= 3

Then A can paint the house in 3 days and B in 2(3)= 6days.

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Answer:

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Step-by-step explanation:

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Read 2 more answers

Plssssss help number 18 and 20​

Plssssss help number 18 and 20