Can someone help me plz

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

6 0

Answer:

33.5883333333

Step-by-step explanation:

If you divided the day and 15 minutes by the speed you would get 33.5883333333

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Somebody please help me with this !!!

Damm [24]

Step-by-step explanation:

Case 1 :

$3x + 2y = 6 \\ 4x + y = 2 \\ 3x = 6 - 2y \\ x = \frac{6 - 2y}{3} = \\ x = 2 - \frac{2y}{3} \\ 4(2 - \frac{2y}{3} ) + y = 2 \\ 8 - \frac{8y}{3} + y = 2 \\ - 1.67y = - 6 \\$

$y = 3.59 \\ x = 2 - \frac{2y}{3} \\x = 2 - \frac{2 \times 3.59}{3} \\ \\ x = - 0.39$

Case 2 :

$3x + 2y = 6 \\ 4x - y = 2 \\ x = 2 - \frac{2y}{3} \\ 4(2 - \frac{2y}{3} ) - y = 2 \\ 8 - \frac{8y}{3} - y = 2 \\ - 3.67y= - 6 \\ y = 1.63$

$x = 2 - \frac{2y}{3} \\ x = 2 - \frac{2 \times 1.63}{3} \\ x = 2 - 1.087 \\ x = 0.913$

7 0

3 years ago

Find an equation of variation in which y varies directly as x and y = 0.4 when x = 0.5. then find the value of y when x=15

Advocard [28]

If x= 0.5 y=0.4
then x=15 y=0.4*15/0.5 = 12

6 0

4 years ago

A company receives shipments of a component used in the manufacture of a component for a high-end acoustic speaker system. When

Tanya [424]

Answer:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

Step-by-step explanation:

For this case we have a sample size of n = 250 units and in this sample they found that 24 units failed one or more of the tests.

We are interested in the proportion of units that fail to meet the company's specifications, and we can estimate this with:

$\hat p = \frac{24}{250}= 0.096$

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 confidence interval for a proportion is given by this formula

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

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.